First let $T$ be the symmetry point of $C$ wrt $PQ$ It's well-known that $\dfrac{CP}{CQ}=\dfrac{DP}{DQ}$, hence $\dfrac{TP}{TQ}=\dfrac{DP}{DQ}$. Plus $\angle PTQ=\angle PCQ=180^\circ-\angle CPQ-\angle CQP=180^\circ-\angle CDP-\angle CDQ=180^\circ-\angle PDQ$, we know $TPDQ$ is a harmonic quadrilateral. From $ADBT$ is a parallelogram, we have $TMD$ are conllinear, and $M$ is midpoint of $TD$. Hence $\triangle PMD\sim\triangle DMQ$, which implies $\angle PMD=\angle DMQ$.

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Here is a similar solution. Let $C'$ be the reflection of $C$ over $B.$ Then, $M$ is the midpoint of $DC'$ and $AB$. Let $R = \ell \cap DC$. Let $X_1 = DC' \cap \ell$ and $X_2$ be on $\ell$ with $\angle X_2MX_1 = 90.$ Observe that $X_2M$ is the perpendicular bisector of $C'D$ and $X_2B$ is the perpendicular bisector of $CC'$, from which it follows that $X_2$ is the circumcenter of $\triangle CC'D.$ Therefore, $\angle CX_2B = \frac12 \angle CPC' = \angle CDC' = \angle CDX_1,$ and so hence $CDX_1X_2$ is cyclic. Therefore, by Power of the Point, we've that $RP^2 = RQ^2 = RD \cdot RC = RX_1 \cdot RX_2,$ which is well-known to imply that $(P, Q; X_1, X_2) = -1.$ Therefore, since $\angle X_1MX_2$, we know that $MX_1$ is the internal angle-bisector of $\angle PMQ$, hence implying $\angle DMP = \angle DMQ$ as desired. $\square$