A decision problem which is not known to be in PH but will be in P if P=NP
If an artificial example is really as good as a natural one, then I can indeed provide such an example!

Edit: Furthermore, my example is "somewhat" less of a cheat than the one suggested by Ravi Boppana (where we take L to be the empty language if P = NP and the halting problem otherwise), in that I'll define the language L by giving a finite procedure to decide whether $x \in$ L for any input x.  At no point will deciding whether x&isin;L require solving an "unbounded" mathematical question such as P vs. NP.



Without further ado: let $M_1, M_2, ...$ be an enumeration of polytime Turing machines.  For all $n$, let $M_{t(n)}$ be the lexicographically first $M_i$ that correctly decides 3SAT on all inputs of length $n$ or less.  Then define the language L as follows: for all inputs $x$ of size $n$, $x \in$ L if and only if the Turing machine encoded by $x$ halts in at most $n^{t(n)}$ steps when run on a blank tape.

Claim 1: If P = NP, then L $\in$ P.

Proof: If P = NP, then there's some fixed $M_i$ that solves 3SAT for all inputs; hence $t(n) \le i$ for all $n$. QED

Claim 2: If P $\ne$ NP, then L $\notin$ P.

Proof: If $t(n)$ increases without bound, then we can simply apply the Time Hierarchy Theorem. QED

Now, not only is L not in P assuming P $\ne$ NP: one presumes it wouldn't be in PH (or even PSPACE) either!

Incidentally, I wonder if anyone can improve the above construction, to get a language L which is in P if P = NP, but provably not in PH or PSPACE if P $\ne$ NP?