$DMEP$ is cyclic. Then it's just simple angle chasing.

I heard that this is the most difficult problem on the test, by far.

Since $AM$ is the angle bisector of $\angle BAC$ and $P$ is the midpoint of arc $BC$, it is obvious that $P$ is located at the second intersection of $AM$ and $\ell$. Case $1$, If we know that $\angle B = 90^{\circ}$ By angle chasing, $\angle DEM = \frac{180^{\circ}-\angle DME}{2} = 90^{\circ}-\frac{\angle DME}{2} = 90^{\circ}-\angle DBE$ $= 90^{\circ}-(180^{\circ}-\angle DPC) = 90^{\circ}-(180^{\circ}-(\angle DPM + 90^{\circ}))$ $= 90^{\circ}-(90^{\circ}-\angle DPM) = \angle DPM$ Hence, $DPEM$ is a cyclic quadrilateral. Since $DM = ME$ and $DPEM$ is a cyclic quadrilateral, it is obvious that $\angle DPM = \angle MPE \implies AP$ is the angle bisector of $\angle DPE$. Case $2$, If we know that $AP$ is the angle bisector of $\angle DPE$ That means, $DPEM$ is a cyclic quadrilateral. By angle chasing, $\angle ABC = \angle APC = 180^{\circ}-(\angle DBE + \angle DPM) = 180^{\circ}-(\frac{\angle DME}{2} + \angle DEM)$ $= 180^{\circ}-(\frac{180^{\circ}-\angle MDE-\angle DEM}{2} + \angle DEM) = 180^{\circ}-(\frac{180^{\circ}-2\angle DEM}{2} + \angle DEM)$ $= 180^{\circ}-(90^{\circ}-\angle DEM + \angle DEM) = 180^{\circ}-90^{\circ} = 90^{\circ}$, as desired. Hence, we have proven that $AP$ is the angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.

For case 2, can you explain why $DPEM$ cyclic? I think thats doesn't work if $DP=PE$.

@above I think I fake solved case 2. Do you have a solution?

The following picture matches the criteria for case 2, but why isn't $\angle ABC = 90^{\circ}$

Did I miss any given information?

I also didn't find a solution if $DP=PE$. I also tried using geogebra with $15$ digit accuracy, but couldn't find the correct construction .-.

This problem only works for case 1 I wonder what the official solution says for case 2

Here's a fixed version of this problem. Let the circumcircle of $\triangle{ABC}$ be $\ell$. There exists a point $M$ in $\triangle{ABC}$ such that $AM$ is the angle bisector of $\angle{BAC}$. The circle centred at $M$ with radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively ($B \neq D, B\neq E$). Let $P$ be the midpoint of arc $BC$ in $\ell$ not containing $A$. Prove that if $\triangle{DPE}$ is not isosceles, line $AP$ acts as the angle bisector of $\angle{DPE}$ if and only if $\angle{B} = 90^{\circ}$.