Complexity of "given a graph $G$ with vertex $v$, is there a maximum clique containing $v$"?
I do not recall having seen this precise problem before, but it does have a Cook reduction (polynomial-time Turing reduction) to $k$-Clique:


  Let $Y$ be an oracle for $k$-Clique. On input $(G,v)$:
  
  
  Determine the maximum clique size $\omega(G)$ via at most $log_{2}(n)$ calls to $Y$ using binary search.
  Create $G'$ by removing all vertices $u \neq v$ where $uv \notin E(G)$ (i.e. $V(G') = N[v]$).
  Determine $\omega(G')$ using $Y$ and binary search.
  If $\omega(G') = \omega(G)$ answer YES, otherwise answer NO.
  


This puts it in $\Delta_{2}^{P} = P^{NP} = P^{SAT}$, a refinement of your observation.

As there are $O(\log{}n)$ calls to the oracle, we can squash containment a little further; the problem is in $\Delta_{2}^{P}[O(\log{}n)] = P^{NP}[O(\log{}n)] = P^{SAT}[O(\log{}n)]$.

[Thanks here to Ricky Demer] If $P=NP$, then $P^{NP} = P^{P} = P$, so you are correct about the polynomial time solvability in this case (contrary to my original, now edited, musings).

So perhaps the problem is $P^{NP}[O(\log{}n)]$-complete, it shouldn't be (as Dominik observed in his answer) $P^{NP}$-complete. I have tried to use the results of Theorem 3.5 from Krentel's paper to construct a reduction, but unless $P^{NP}$ is low for itself (which I don't think it is, but don't know), this has been without luck. (If it is low for itself, then I have a marvellous proof, sadly there are no margins here to put it in.)