Why doesn't P=NP imply P=AP (i.e. P=PSPACE)? It is well known that if $\mathbf{P}=\mathbf{NP}$ then the polynomial hierarchy collapses and $\mathbf{P}=\mathbf{PH}$.

This can easily be understood inductively using oracle machines.
The question is - why can't we continue the inductive process beyond a constant level of alternations and prove $\mathbf{P}=\mathbf{AltTime}(n^{O(1)})$ (aka $\mathbf{AP}=\mathbf{PSPACE}$)?

I am looking for an intuitive answer.
The proof for $\mathbf{P}=\mathbf{AltTime}(O(1))$ ($=\mathbf{PH}$) is an induction using $\mathbf{P}=\mathbf{NP}$. The induction shows that for any natural number $k$, $\mathbf{P}=\mathbf{AltTime}(k)$ (and $\mathbf{AltTime}(O(1))$ is just their union). 

The induction does not work when the number of alternation can change with the input size (i.e. when the number of possible alternations of the machine is not a number but a function of the input size, i.e. we are not showing that an execution of the machine on a single input can be reduced to no alternation, we are showing that the executions of the machine on all inputs can be "uniformly" reduced to no alternation).

Let's look at a similar but simpler statement. We want to show that the identity function $id(n)=n$ eventually dominates all constant functions ($f \ll g$ iff for all but finitely many $n$ $f(n) \leq g(n)$). It can be proven say by induction. For all $k$, $k \ll n$ (i.e. $f_k \ll id$ where $f_k(n)=k$), but we don't have this for non-constant functions like $n^2$, $n^2 \not \ll n$.