Since $\angle EFM = \angle AFB = \angle DCA = \angle ECM$, the points $M$, $C$, $F$, $E$ are concyclic. Using the concyclic points, we obtain
$$\angle QME = \angle FME = \angle FCE = \angle FCD = \angle FBD = \angle QBD.$$
This implies $EM \parallel DB$, and hence $\frac{DE}{EQ} = \frac{BM}{MQ}$. Also, as $\angle MCQ = \angle MCB$, we have $\frac{BM}{MQ} = \frac{BC}{CQ}$. Combining these, we obtain $\frac{DE}{EQ} = \frac{BC}{CQ}$. It follows that $EQ = CQ$. Now, as $M$ and $Q$ are midpoints of $CP$ and $CE$, we know that $PE \parallel MQ$ as desired.