From $\triangle BMC \sim \triangle PBC$ follows $\frac{BM}{PB} = \frac{BC}{PC}$, hence $CP = 2BP$, so $[BP] = [PN] = [NC]$.

Then $\Delta NCD \equiv \Delta PBC$ (SAS) leads to $\widehat{DNC} = \widehat{BPC}$, that is $DN \perp CP$, so $[DN]$ is a median and an altitude in triangle $DPC$. Therefore triangle $DPC$ is isosceles, with $[DP] = [DC]$.

Denote $E$ the common point of the straight lines $CM$ and $AD$. Then $A$ is the midpoint of $[DE]$, hence $[NA]$ is a median in the right triangle $NDE$. This leads to $[NA] = [AD]$, hence triangle $ADN$ is isosceles. Since $AQ$ is a bisector, it is also an altitude, therefore $AQ \perp DN$. From $DN \perp CM$ follows $AQ \parallel MN$ (1).

Isosceles triangles $ANE$ and $DPC$ yield $\widehat{ANP}+\widehat{DPN} = \widehat{DEC}+\widehat{DCE} = 90^\circ$, whence $DP \perp AN$. Since $AQ \perp DN$, the point $Q$ is the orthocenter of the triangle $ADN$, so $NQ \perp AD$.

This shows that $NQ \parallel AM$ and, using (1), we get that $AMNQ$ is a parallelogram.
So the segments $[AM]$ and $[NQ]$ are parallel and congruent, hence $[BM]$ and $[NQ]$ are parallel and congruent, therefore $BMQN$ is a parallelogram.