This one is going to be hard to explain in ordinary terms, and if tutors such as Arnold comes in, he might be able to offer better advice than me. After all, I'm not specialised in the sciences, but I will still try to explain it to you.

By definition, potential difference (p.d.) is the work done to drive a unit charge through a particular electrical component. Work done is, in a way, like energy (after all, energy is defined as the capacity to do work).

Let's divert a little and talk about gravitational potential energy (GPE) for now. Supposing you have an object on top of a building. It falls and reaches ground level. What is its GPE at ground level?

You might think it would be zero, which at your level will likely be the case. But let's think for a moment. What happens if...the ball could fall further beyond ground level? How can it fall further if it has zero potential energy at ground level?

The thing is, gravitational potential energy works on a "relative" basis. In other words, it works based on height levels. The difference in height levels will then lead to this difference in potential energy levels.

We are actually able to set our reference heights arbitrarily (for example, ground level can be zero potential energy, or even 30 m below ground level can be set to zero). No matter the choice we make, our calculated difference in potential energy will remain the same in value.

You can see that we can actually kind of set reference points for gravitational potential energy (though this only becomes clearer at higher level physics).

For electrical potential energy, they will work in a similar manner (apart from the obvious fact that this potential energy is based on electrical factors). Let's now put our focus back to the question.

Conventional current is taken to be the direction of current flow from the positive terminal of the battery to the negative terminal of the battery.

Electric potential is a measure of energy levels. The difference in energy levels is then called the difference in potential, or in short the potential difference (as I defined in potential difference earlier).

You can largely think of this as energy. But I can set the energy levels arbitrarily the way I want it to be.

I will set the positive terminal to be at a potential level of 6 V while the negative terminal to be at a potential level of 0 V (hence, the "potential difference" of 6 V across both terminals of the battery, better known as the electromotive force).

I like to set my negative terminal to be 0 V because I think of the current flowing from the positive terminal to the negative terminal as a person's journey from day to night. He starts out full of energy...and ends with zero energy. Ok, not exactly zero, but you get the gist of what I mean.

An important rule to know is that all points on the same piece of wire, uninterrupted by any resistor, lamp or similar, are on the same potential level. We assume that "no energy" is lost as long as we do not attempt to cross a barrier in our journey (here, a resistor).

I'll use option A to illustrate my point.

The current starts out its journey on a high potential level of 6 V. At all points towards any of the 10 ohm resistors, its potential level is always 6 V.

The current always ends its journey on a low potential level of 0 V. So, at all points past the 5 ohm resistors, its potential level is 0 V.

At P and Q, however, the energy level is neither 6 V nor 0 V. This is because for a "two-obstacle journey", we would need to conserve some energy for each obstacle.

I'll take the branch containing "P" in it for now. The 10 ohm resistor has twice as much resistance as the 5 ohm resistor. Because they are connected in a "mini-series" arrangement with each other, the current in that mini-branch is the same throughout.

By the formula V = IR, the 10 ohm resistor will "consume" twice as much "potential difference" as the 5 ohm resistor.

Considering the fact that the overall electromotive force is 6 V, the 10 ohm resistor must "consume" 4 V of potential, while the 5 ohm resistor must "consume" 2 V of potential.

A drop in 4 V potential across the 10 ohm resistor means that the current, having an initial potential level of 6 V, must lose 4 V of its potential to end up with 2 V of potential at point P. This further drops by 2 V across the 5 ohm resistor to end up with 0 V at the end, as expected.

So, the potential level at P is 2 V.

Working out the branch containing "Q" in it, point Q is also at a potential level of 2 V.

Since the potential levels at P and Q are both 2 V, there is no "difference" between them. Hence, there is no potential difference between points P and Q.

In option B, it is similar in idea.

Again, the drop in potential level is 2 V across the 5 ohm resistor and 4 ohm across the 10 ohm resistor.

Points P and Q must therefore be at the same potential level of 4 V (due to the drop in the initial 6 V potential level by a figure of 2 V).

There is no potential difference between P and Q.

Option C is trickier, but I can assure you that there is no potential difference here.

For the branch containing "P", both resistors are of value 10 ohms.

Both resistors will consume the same amount of potential level as they attempt to split the potentials according to their ohm values.

The drop across each 10 ohm resistor is 3 V.

So, point P is at a potential level of 3 V (a drop of the initial 6 V potential level by a figure of 3 V).

For the branch containing "Q", both resistors are of value 5 ohms.

Both resistors will consume the same amount of potential level as they attempt to split the potentials according to their ohm values.

The drop across each 5 ohm resistor is 3 V.

So, point Q is at a potential level of 3 V (a drop of the initial 6 V potential level by a figure of 3 V).

There is no potential difference between points P and Q.

Note that the currents in branch P and branch Q are NOT the same - but their potential levels are.

For option D, the arrangement of the resistors plays a role in the potential value.

Again, because each branch contains a 5 ohm resistor and a 10 ohm resistor, the drop across the 5 ohm resistor is 2 V while the drop across the 10 ohm resistor is 4 V.

Branch P faces the 5 ohm resistor first. The initial potential drop must be 2 V, hence reducing the potential level at P to 4 V.

Branch Q faces the 10 ohm resistor first. The initial potential drop must be 4 V, hence reducing the potential level at Q to 2 V.

Because the potential level at P is 4 V while the potential level at Q is 2 V, there is a difference in value between points P and Q. Hence, there is a potential difference (of 2 V) between points P and Q.