Lets take a boolean function f(A,B)=A+B of 2 boolen variables.

A B A+B

0 0 0

0 1 1

1 0 1

1 1 1

Suppose we have to find out any A,B combination for which f(A,B) gives value 1, to do so we might have check all (2^{n}=2^{2}=4) conbinations for which f(A,B) may give value 1.

But for a given combination(A,B) if we have to check whether the functional output is 1 or not we can easily do that in polynomial time. Hence option (B)

So as we can't find out solution of such given problem in polynomial time but given a solution we can always verify in polynomial time,so this problem falls under NP category.