Solution:

Let lines $BE$ and $AC$ meet at $R$. It suffices to show that the points $P$, $M$ and $R$ are collinear. Since $AP$ is tangent to the circumcircle of $ABC$ and $PNA M$ is cyclic, we have $\angle PNM = \angle PAB = \angle ACB$ and $\angle BAC = \angle MPN$. Thus, triangles $MPN$ and $BAC$ are similar with $\frac{MP}{MN} = \frac{AB}{BC}$. Also, since $DM$ and $PE$ are parallel and $DPEM$ is cyclic, we get $\angle PMD = \angle MPE = \angle ABR$, so that triangles $PMD$ and $ABR$ are similar with $\frac{PD}{PM} = \frac{AR}{AB}$.

Now, observe that $\angle PER = \angle PEB = 180^\circ - \angle PMB = 180^\circ - \angle PNR$, so $PERN$ is cyclic and $\angle CRB = \angle DPE = \angle NDM$. We see that triangles $CRB$ and $NDM$ are similar with $\frac{CB}{CR} = \frac{MN}{ND}$. With $QC = PB$, we have $QB = PC$ and Thales' theorem gives $\frac{PD}{DN} = \frac{PB}{BQ} = \frac{PB}{PC}$. Thus, applying Menelaus' theorem on triangle $PAC$, we obtain

$$\frac{CR}{AR} \cdot \frac{AM}{MB} \cdot \frac{BP}{PC} = \frac{CB \cdot ND}{MN} \cdot \frac{PM}{PD \cdot AB} \cdot \frac{PD}{DN} = \frac{PM}{MN} \cdot \frac{BC}{AB} = 1$$

implying that $P$, $M$ and $R$ are collinear as desired.