Without loss of generality, we can assume that $ CA<CB $ so that $ \odot(BMP) $ is tangent to $ AC $ beyond $ C $ at $ K $. Then by Power of a point we get $ AK=\frac{AB}{\sqrt2} $ and $ CK^2=CP.CB $. But note that $ CP.CB $ is also the power of $ C $ wrt the circle on diameter $ AB $, hence we get $ CP.CB=CM^2-\frac{AB^2}{4} $. Using the median theorem in a triangle, $ CM^2=\frac{2CA^2+2CB^2-AB^2}{4} $ and thus we have $ CP.CB=\frac{CA^2+CB^2-AB^2}{2} \implies CK=\sqrt{\frac{CA^2+CB^2-AB^2}{2}} $. We then have $ AC=AK-CK \implies $ $ \frac{AB}{\sqrt2}-AC $ $ =\sqrt{\frac{CA^2+CB^2-AB^2}{2}} \implies $ $ \frac{AB^2}{2}+AC^2-\sqrt2AB \cdot AC=\frac{CA^2+CB^2-AB^2}{2} \implies 2AB^2-2\sqrt2AC\cdot AB+AC^2-BC^2=0 $ Now using the quadratic formula, the length $ AB $ is given by $ \frac{2 \sqrt2 AC + \sqrt{8AC^2-8(AC^2-BC^2)} }{4} $ (Note that $ AB $ is positive so we do not consider the negative root). This gives $ AB=\frac{CA+CB}{\sqrt2} $. Let $ L $ be the point on $ BC $ such that $ BL=AK $. If $ L $ lies beyond $ C $(or $ L=C $) on $ BC $ then we get $ \sqrt2AB=CA+CB<AK+BL=\sqrt2AB $ a contradiction. So $ L $ is surely between $ B,C $. Thus we get $ AK+BL=\sqrt2AB=AC+CB \implies CK=CL $. Obviously circle $ (AML) $ is tangent to $ BC $. if it cuts $ CA $ again at $ Q' $ then we obtain $ CQ'.CA=CL^2=CK^2=CP.CB $ hence, $ Q' $ is on $ (ABP) \implies Q' $ is the foot of altitude from $ B $ thus $ Q'=Q $ and $ \odot(AMQ) $ is tangent to $ BC $ as desired.

Let $MD$ cut $BC$ at $E$, easy to see that $\angle EPD = \angle DMB$ and $\angle EDP = \angle PBM = \angle MPB = \angle MDB, \Delta EDP\sim \Delta BMD$, thus $\angle DEC = \angle DBM = \angle CDE$, or $EC = DC$. Using Menelaus in $\Delta ABC$ with line $MED$ easy to obtain $AD = EB$. Now notice that $BE^{2} = AD^{2} = AM. AB = BM. BA$, this say that the circle $(AME)$ is tangent to $BE$. But as $CQ. CA = CP. CB = CD^{2} = CE^{2}, Q$ is a point on this circle and we have q.e.d.

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