Given an NFA A and a regular expression B, is the problem of determining L(A) = L(B) decidable?
Yes. Transform both into DFAs $M_1$ and $M_2$; from them construct DFAs for $L_1 \cap \overline{L_2}$ and $\overline{L_1} \cap L_2$, and check that neither accepts anything.

As noted in the comments, the NFA to DFA construction can give an exponential increase in number of states, same for regular expression to DFA, so this doesn't prove the problem is in NP. Luckily you weren't asked about that ;-)