How to prove $P^{Halt} = PSPACE^{Halt}$
Using $n$ calls to the halting oracle and time $O(n^2)$, you can compute the first $n$ bits of the Chaitin's constant.  Using the $n$ bits of the Chaitin's constant and unbounded time, all queries to the halting oracle of length approximately $≤n$ (depending on the representation) can be eliminated.  Finally, the result can be obtained in bounded time by querying the halting oracle.

Additional Details: In the relativization of $\mathrm{PSPACE}$ here, the oracle tape counts against space usage, so a $\mathrm{PSPACE}^\mathrm{Halt}$ machine can only use queries of length $n^{O(1)}$ (though the number of the queries can be exponential).  The digits of the Chaitin's constant can be used to get the number of machines of length $≤n$ that halt, and by simulating all these machines until the right number halts, we can find out exactly which of the (roughly $2^n$) machines halt.  The digits of the Chaitin's constant, or just the number of the halting machines, can be computed in $n^{O(1)}$ queries by bisection.