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 (2n=22=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.