Let $F$ be the intersection of $PB$ and $(CDE)$, $H$ the orthocener of $ABC$ and $T$ the intersection of $(CDE)$ and $(AHB)$. Claim $P=T$ By angle chasing we have, $$\angle BAT=\angle BHT=180-\angle EHT= \angle ACT$$So $AB$ is tangent with $APC$ and similarly with $BPC$. So it is well known that $CT$ bisects $AB$ and hence the claim. Now by Reim's theorem in $(CBE)$ and $(BPHA)$ we have that $FD$ is parallel to $AB$ so, $$\angle AQP= \angle AQC= \angle B= \angle FDC= \angle FPC = \angle QPB$$Hence $AQ$ is parallel to $BP$ and combined with the fact that $M$ is the midpoint of $AB$ means that $AQPB$ is a parallelogram and therefore $MP=MQ$

Lemma 1: $MD$ and $ME$ are tangents of $(CDE)$ Proof: $\angle{ACB}= 90 - \angle{EBC} =90- \frac{\angle{EMD}}{2} = \angle{MDE}$ Then, by PoP: $ MP \times MC = MD^2 = MB^2$ $\Rightarrow$ $MP/MB=MB/MC$ $\Rightarrow$ $\triangle{MBC} \sim \triangle{MPB}$ $\Rightarrow$ $\angle{MPB}=\angle{ABC}=\angle{AQC}$ So, $AQ$ and $BP$ are parallel. Then, since $AM=MB$, $AQBP$ is a parallelogram and $MQ=MP$.