Note that $$\angle PAB = \angle DAB - \angle DAP = \angle DAC - \angle DMP = \angle CDM - \angle DMB$$$$= 180^{\circ}-\angle BDM - \angle DMB = \angle DBM = \angle DBP.$$This means that $(ABP)$ is tangent to line $BC$ so if we let $X = AP \cap BC$ we have $XB^2 = XP \cdot XA = XD^2$ meaning $XB=XD$ so $AP$ is indeed a median on triangle $ABD$.

First step of the proof is to prove that $\omega $ is tangent to the circumcircle of $ABC$, which is easy by Archimede's Lemma. Let's say $AP$ meets $BD$ at $N$. By P.O.P we have $ND^2=NP\cdot NA$. So if we want to prove that $NB=ND$, we shall prove that $NB^2=NP\cdot NA$, equivalent with proving that $NPB$ and $NAB$ are similar, so it's enough to prove that $\angle APM=\angle ABC$, which is again an easy angle chasing with the tangents( taking any point X on the common tangent of $\omega$ such that $X$ and $B$ aren't the same side by $AC$ we have $\angle APM=\angle XAM=\angle ABC$). Thus, the proof is finished so $ND=NB$.

$\angle DBP=180-\angle BPD-\angle BDP=\angle MPD-\angle DAP=\angle DAB-\angle DAP=\angle BAP$ So $BC$ is tangent to $(BAP)$ and it is well known that this leads to the proof.

dame dame

The angle chase here is quite motivated as we know by a well known lemma ,(In EGMO chapter 2),about when does P lie on the median ,Given the conditions, it remains to prove that $(BAP)$ is tangent to $BC$ which is true by an angle chase.