Let $R,S$ be the midpoints of the diagonals $AC,BD.$ Since $BC=AD,$ then $NR=NS$ $\Longrightarrow$ Parallelogram $NRMS$ is a rhombus $\Longrightarrow$ $NM$ bisects $\angle RNS$ internally. Let $U \equiv AD \cap BC.$ Since $\angle CUD$ and $\angle RNS$ have corresponding parallel sides, it follows that their angle bisectors are parallel as well. By Menelaus' theorem for $\triangle UCD$ cut by $\overline{QPN},$ keeping in mind that $UP=UQ$ and $CN=ND,$ we get $\frac{QU}{QC} \cdot \frac{CN}{ND} \cdot \frac{DP}{PU}=1 \Longrightarrow \ QC=DP.$

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=377297 then use sine law.

It is easy to see that $MN$ is parallel to the bisector of the angle $\widehat {(BC, AD)}$, hence the triangles $\triangle MBQ$ and $\triangle AMP$ are pseudo-similar and our problem is solved then. Best regards, sunken rock

By Law of Sines on $\triangle MPA$ and $\triangle QMB$, $$\frac{AM}{\sin{\angle MPA}}= \frac{AP}{\sin{\angle AMP}}$$and $$\frac{BM}{\sin{\angle MQB}}= \frac{BQ}{\sin{\angle BMQ}}$$Since, $\sin{\angle AMP}=\sin{\angle BMQ}$, we get that $$\frac{\sin{\angle MQB}}{\sin{\angle MPA}}= \frac{AP}{BQ}.$$ By Law of Sines on $\triangle NPD$ and $\triangle QNC$, $$\frac{DN}{\sin{\angle NPD}}= \frac{DP}{\sin{\angle DNP}}$$and $$\frac{CN}{\sin{\angle CQN}}= \frac{CQ}{\sin{\angle CNQ}}$$Since, $\sin{\angle DNP}=\sin{\angle CNQ}$, we get that $$\frac{\sin{\angle CQN}}{\sin{\angle NPD}}= \frac{DP}{CQ}.$$Because, $\angle MQB=\angle CQN$ and $\angle MPA=\angle NPD$, we get that $$\frac{AP}{BQ}= \frac{DP}{CQ}\Longleftrightarrow \frac{AD+DP}{DP}= \frac{BC+CQ}{CQ}\Longleftrightarrow \frac{AD}{DP}= \frac{BC}{CQ}.$$Since $BC=AD$, we get that $CQ=DP$. $\quad \square$

Let $S$ be the center of spiral similarity that sends $A \rightarrow B, D \rightarrow C$, $X=AD \cap BC$. Therefore, it's also the center of spiral similarity that sends $M \rightarrow N$ and $BS=AS$. Hence, $\square ABSX, \square BMQS$ are cyclic quadrilaterals, also $\square PXQS$. Hence, there is a spiral similarity center at $S$ that sends $BQ \rightarrow AP$. But $BS=AS \implies BQ = AP \implies CQ = DP$ .