4/52 + 12/52 = 4/13
Out of 13 spades, only 12 spades will not have 7. The question is asking for spades OR a 7.

Because, in doing 4/52 + 13/52, you were actually double counting the 7 of spades.

We need either a spade or a 7. As long as either condition is satisfied, we can consider that card.

7 ♥, 7♦,7♣ satisfy the second condition.

2,3,4,5,6,8,9,10,J,Q,K,A ♠ satisfy the first condition.

7 ♠ satisfies both. But you must be careful not to double count it since there is only 1 of it.

So the answer is 16/52 = 4/13


If the question asked for 7 only, then B (4/52 = 1/13) would be the answer.

If the question asked for both 7 and a spade, then A (1/52) is correct.

Event A: getting a spade : 13/52
Event B: getting a 7: 4/52
Note- there is a likelihood of 1/52 chance of getting both a spade and a 7 in event A. So if you sum both events A and B, it would be double inclusion of the event B of getting a 7 which also happens to be a spade.
Hence, P(A union B) = P(A) + P(B) - P(A intersect B) = 13/52 + 4/52 -1/52=4/13