It's difficult to draw a accurate diagram. I use Geogebra and I find that $\angle DME=90^\circ$ when $\angle BPD=\angle CME$.

Define $\Gamma_B \equiv \odot(D, DB)$ and $\Gamma_C \equiv \odot(E, EC)$ and let $Q$ be the second intersection of $\Gamma_B$ and $\Gamma_C.$ Let $X$ be the reflection of $C$ in $E.$ We will show that $\measuredangle DPB = \measuredangle CME \iff \measuredangle BQC = 90^{\circ}.$ Analagously one may obtain $\measuredangle EPC = \measuredangle BMD \iff \measuredangle BQC = 90^{\circ}$ and the desired result will follow. Note that $DE \perp PQ \implies BC \perp PQ.$ Then since $D$ is the circumcenter of $\triangle PQB$ we have $\measuredangle DPB = 90^{\circ} - \measuredangle BQP = \measuredangle CBQ.$ On the other hand, $\measuredangle CME = \measuredangle CBX$ because $ME$ is the C-midline of $\triangle BXC.$ Therefore, $\measuredangle DPB = \measuredangle CME \iff \measuredangle CBQ = \measuredangle CBX \iff B \in QX.$ But since $\overline{CX}$ is a diameter of $\Gamma_C$, we have $\measuredangle CQX = 90^{\circ}.$ Hence, $\measuredangle DPB = \measuredangle CME \iff \measuredangle BQC = 90^{\circ}$, as desired. $\square$

Very nice problem but I think it is a bit too hard for CHKMO, it took me around $3-4$ hours to found the solution(mainly because techniques in my regular toolbox such as trigonometry, spiral similarity and projective geometry are useless in this problem.) Also you need to construct a lot of extra lines in this problem, so I think it is very hard to solve it under contest condition with only $3$ hours to spent. Anyways, this is the solution. Let $B',C'$ be the reflection of $B,C$ in $EM,DM$ respectively, since $MB=MC=MB'=MC'$, $B,C,B',C'$ lies on a circle with diameter $BC$. Let $\Gamma_d$ and $\Gamma_e$ be the circles cenetred at $D$ with radius $DB$ and the circle centreed at $E$ with radius $EC$ respectively. Suppose they intersect at two points $P$ and $P'$. Now by radical axis theorem on $\Gamma_d,\Gamma_e$ and $(BCC'B')$ we have $BC',CB',P'P$ are concurrent at a point $H$. Moreover, $PP'\perp DE$, hence $PP'\perp BC$. Now since $\angle BPD=\angle CME$, $$\angle BP'P=90^{\circ}-\angle BPD=90^{\circ}-\angle CME=\angle C'CB$$Therefore, $BP'\perp CC'$ and $H$ is the orthocenter of $\triangle BP'C$, hence $\angle CP'P=\angle B'BC$ This implies $$\angle CPE=90^{\circ}-\angle PP'C=90^{\circ}-\angle B'BC=\angle BMD$$