Let $X$ and $Y$ be the intersection points of $AM$ and $BM$ with $PD$ and $PC$ respectively. Since $ABCM D$ is cyclic and $CM = MD$, we have
$$\angle XAD = \angle MAD = \angle CBM = \angle CBY.$$
Combining this with $\angle ADX = \angle YCB$, we get $\angle DXA = \angle BYC$, and so $\angle PXM = \angle MYP$. Moreover, $\angle YPX = \angle CPD = \angle ADB = \angle AMB$. The quadrilateral $MXPY$ therefore has equal opposite angles and so is a parallelogram.

Let $R$ and $S$ be the intersection points of $AM$ and $BM$ with $BC$ and $AD$ respectively. Due to $AM \parallel PC$ and $BM \parallel PD$, we have $\angle ASB = \angle ADP = \angle PCB = \angle ARB$ and so the quadrilateral $ABRS$ is cyclic. We then have $\angle SRB = 180^{\circ} - \angle BAS = \angle DCB$ and so $SR \parallel CD$. In triangles $PCD$ and $MRS$, the corresponding sides are parallel so they are homothetic meaning lines $DS$, $PM$, $CR$ concur at the centre of this homothety.

Let $AD$ and $BC$ meet at $T$. Denote by $p_{a}, p_{b}, m_{a}$ and $m_{b}$ the distances between line $TA$ and $P$, $TB$ and $P$, $TA$ and $M$ and between $TB$ and $M$ respectively. Our goal is to prove $p_{a} : p_{b} = m_{a} : m_{b}$ which is equivalent to the collinearity of $T$, $P$ and $M$.

Let $\angle BAC = \angle BDC = \alpha$, $\angle DBA = \angle DCA = \beta$, $\angle ADB = \angle AMB = \angle ACB = \angle CPD = \mu$, $\angle ADP = \angle PCB = \nu$ and $\angle MAD = \angle CAM = \angle MBD = \angle CBM = \chi$.
From $\angle ADP = \angle PCB = \nu$ and $\angle MAD = \angle CBM = \chi$ we get
$$\frac{p_{a}}{p_{b}} = \frac{PD \cdot \sin \nu}{PC \cdot \sin \nu} = \frac{PD}{PC} \quad \text{and} \quad \frac{m_{a}}{m_{b}} = \frac{MA \cdot \sin \chi}{MB \cdot \sin \chi} = \frac{MA}{MB}.$$
Hence $p_{a} : p_{b} = m_{a} : m_{b}$ is equivalent to $PD : PC = MA : MB$, and since $\angle CPD = \angle AMB = \mu$, this means we have to show that triangles $PDC$ and $MAB$ are similar.
In triangle $PDC$ we have
$$\begin{aligned} & \angle PDC + \angle DCP = 180^{\circ} - \angle CPD = 180^{\circ} - \mu, \\ & \angle PDC - \angle DCP = (\alpha + \mu - \nu) - (\beta + \mu - \nu) = \alpha - \beta. \end{aligned}$$
Similarly, in triangle $MAB$ we have
$$\begin{aligned} & \angle BAM + \angle MBA = 180^{\circ} - \angle AMB = 180^{\circ} - \mu, \\ & \angle BAM - \angle MBA = (\alpha + \chi) - (\beta + \chi) = \alpha - \beta. \end{aligned}$$
Therefore, ($\angle BAM$, $\angle MBA$) and ($\angle PDC$, $\angle DCP$) satisfy the same system of linear equations. The common solution is
$$\angle BAM = \angle PDC = \frac{180^{\circ} - \mu + \alpha - \beta}{2} \text{ and } \angle MBA = \angle DCP = \frac{180^{\circ} - \mu - \alpha + \beta}{2}.$$
Hence triangles $PDC$ and $MAB$ have equal angles and so are similar. This completes the proof.