Note that $\angle BPD = \angle CME = \angle DEM$. So we can find a point $F$ on the extension of $EM$ such that $\triangle DPB \sim \triangle DEF$. By spiral similarity and since

$DB = DP$, we have $\triangle DFB \cong \triangle DEP$. As $EC = EP = FB$ and $CM = BM$, by the sine law, either $\angle FEC = \angle EFB$ or $\angle FEC + \angle EFB = 180^\circ$.
If $\angle FEC + \angle EFB = 180^\circ$, then
$$\begin{align*} \angle AED &= 180^\circ - \angle FEC - \angle DEM \\ &= \angle EFB - \angle DEM \\ &= \angle DFB + \angle EFD - \angle DEM \\ &= \angle DEP. \end{align*}$$
Together with $\angle AEP = 2\angle ECP$, we easily find that $DE \parallel PC$, contradicting the fact that $P$ does not lie on $BC$.

Hence, we must have $\angle FEC = \angle EFB$. This yields $EC \parallel BF$. Therefore, we find that $EM = FM$. Since $DE = DF$, we have $DM \perp EF$. Now, by swapping the roles of $(B, D)$ and $(C, E)$, and reversing the above arguments, we see that $\angle CPE = \angle BMD$ as desired.