We deploy barycentric co-ordinates and a little thought. Let $ \Gamma$ denote the circle tangent to $ AB$, $ AC$ and $ \Omega$, and $ \Psi$ the incircle. $ P$ arises as the external centre of similitude of $ \Gamma$, $ \Omega$ and also lies on $ \Omega$. Note that a positive homothety through $ A$ takes $ \Psi$ to $ \Gamma$, and the centres of two homotheties and the centre of their composition are collinear. We can thus characterise $ P$ as the intersection of line $ AX_{56}$ with $ \Omega$ ($ P \not= A$), where $ X_{56}$ is the external centre of similitude of the incircle and circumcircle, which has a standard barycentric characterisation in Kimberling's encyclopedia of triangle centres. $ X_{56}(\frac{a^2}{b+c-a},\frac{b^2}{c+a-b},\frac{c^2}{a+b-c})$ So line $ AX_{56}$ can be parametrically represented as $ (\frac{a^2}{b+c-a},\frac{tb^2}{c+a-b},\frac{tc^2}{a+b-c})$. Intersecting with the circumcircle $ a^2yz+b^2zx+c^2xy=0$, after cancelling $ a^2b^2c^2$ and clearing denominators we get \[ t^2(b+c-a) + t(c+a-b+a+b-c)=t(t(b+c-a)+2a)=0\] which is obviously true (since $ 2s=a+b+c$). Since we have $ A \not= P$, $ t\not=0$, so $ t=\frac{-2a}{b+c-a}$, yielding the co-ordinates of $ P$ as \[ P(-a, \frac{2b^2}{c+a-b}, \frac{2c^2}{a+b-c}) = P(-a(s-b)(s-c), b^2(s-c), c^2(s-b))\]. The other two points are rather easier to obtain. The point at infinity intersecting $ x=0$ is $ (0,-1,1)$ so the line through $ A$ parallel to $ BC$ contains $ (1,0,0),(0,-1,1)$, and is therefore $ (1,-t,t)$ ($ t=0,\infty$ will do it). Intersecting with the circumcircle, $ t(b^2-c^2)=a^2t^2$. Again, unless $ b=c$ we have $ M$ distinct from $ A$, so it corresponds to the root $ t=\frac{b^2-c^2}{a^2}$. Thus \[ M(a^2, c^2-b^2, b^2-c^2)\]. Finally, $ D(0,s-c,s-b)$ is well known and obvious. To show $ D,M,P$ are collinear it will thus suffice to show that the determinant of the 3x3 matrix with rows equal to these three triples of co-ordinates vanishes. This is given by the expression (taking DMP as the order in my matrix, not that it matters) \[ (s-c)((b^2-c^2)(-a(s-b)(s-c)) - a^2c^2(s-b)) + (s-b)(a^2b^2(s-c)+a(c^2-b^2)(s-b)(s-c)\] This vanishes (cancelling through $ a(s-b)(s-c)$) iff \[ ab^2+(s-b)(c^2-b^2) = ac^2 + (s-c)(b^2-c^2)\] Which is equivalent to $ a(b^2-c^2) = (s-c-s+b)(b^2-c^2)$. And this is clearly true by $ 2s=a+b+c)$, so we are done.

Let the circle which tangent to $ AB,AC$ and the circumcircle be $ w$ . Let $ w$ touching the sides $ AB,AC$ at the points $ Q,R$ respectively . Let $ B',C'$ the midpoints of the arcs $ AC,AB$ which don't contain the points $ B,C$ .The the homothety with center $ P$ that sends $ w$ to the circumcircle. It's easy to see that $ Q$ goes to $ C'$ and $ R$ goes to $ B'$ which means that the lines $ PQ,PR$ are bisectors of the angles $ \angle{APB},\angle{APC}$ respectively .And also from pascal's theorem to the six points $ A,B',C,P,B,C'$ we have that the points $ Q,I,R$ are collinear , where $ I$ is the incenter of $ ABC$ .Now from the bisector theorem we have $ \frac {QB}{QA} = \frac {PA}{PB}$ and $ \frac {RA}{RC} = \frac {PA}{PC}$ and dividing these two we have that $ \frac {PC}{PB} = \frac {RC}{QB}$ (1). From the law of sines at the triangles $ PBD,PDC$ we find that $ \frac {\sin\angle{BPD}}{\sin\angle{DPC}} = \frac {CP\cdot BD}{BP\cdot CD}$ (2) Now we have that since $ \angle{ARI} = 90 - \frac {A}{2}$ and $ \angle{RCI} = \frac {C}{2}$ we will have that $ \angle{RIC} = \frac {\angle{B}}{2}$ so from the law of sines at the triangle $ RIC$ we have that $ RC = IC\cdot\frac {\sin\frac {B}{2}}{\cos\frac {A}{2}}$ and similarly we have $ BQ = IB\cdot\frac {\sin\frac {C}{2}}{\cos\frac {A}{2}}$ So $ \frac {RC}{QB} = \frac {IC}{IB}\cdot\frac {\sin\frac {B}{2}}{\sin\frac {C}{2}}$ But $ \frac {IC}{IB} = \frac {CD}{BD}\cdot\frac {\cos\frac {B}{2}}{\cos\frac {C}{2}}$ so $ \frac {RC}{QB} = \frac {CD}{BD}\cdot\frac {\sin B}{\sin C}$ (3) So the relation (2) with the help of (1) and (3) becomes $ \frac {\sin\angle{BPD}}{\sin\angle{DPC}} = \frac {\sin B}{\sin C}$ so $ \frac {\sin {(B + C - \angle{DPC})}}{\sin\angle{DPC}} = \frac {\sin (B + C - C)}{\sin C}$ But the function $ f(x) = \frac {\sin{(a - x)}}{\sin x}$ is monotonous so $ \angle{BPD} = B$ and $ \angle{DPC} = C$ . Now if we let PD meets the circumcircle at $ M'$ we will have that $ \angle{M'PC} = C$ or $ \angle{M'AC} = C$ so the lines $ AM'$ and $ BC$ are parallel so $ M\equiv M'$ and we are done .

LEMMA Given a triangle $ ABC$, where $ \alpha$ is circumcircle. $ P \in \alpha$, such that $ AP$ is external bisector. $ I$ incenter of $ ABC$. $ \beta$ is the circle internally tangent to $ \alpha$,$ AB$ and $ AC$, $ \alpha$ touches $ \beta$ in $ S$.Then $ P,I,S$ are collinear. In the problem,

Lemma: In a triangle $ABC$ with circumcircle $\Omega$, let $J$ be the midpoint of arc $\widehat{BAC}$ and $I$ be the incenter of triangle $\triangle{ABC}$. If $P$ is the point such that the circle $\omega$ tangent to segments $AB$ and $AC$ and to $\Omega$ is tangent to $\Omega$, then $P$, $I$ and $J$ are collinear. Proof of Lemma. Let $R$ and $S$ denote the midpoints of arcs $\widehat{AB}$ and $\widehat{AC}$ not containing $C$ and $B$, respectively, and let $\omega$ be tangent to sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $\mathcal{H}$ be the homothety sending $\omega$ to $\Omega$. Note that $\mathcal{H}$ has center $P$. It follows that $\mathcal{H}(X)=R$ and $\mathcal{H}(Y)=S$ and hence that $P$, $X$ and $R$ are collinear and that $P$, $Y$ and $S$ are collinear, since $P$ is the center of homothety $\mathcal{H}$. Hence by Pascal's theorem $X$, $I$ and $Y$ are collinear, since $I$ is the intersection of $CR$ and $BS$. Further since $AX$ and $AY$ are tangents to $\omega$, it follows that $AX=AY$ and, since $AI$ bisects $\angle{XAY}$, it follows that $XY$ is perpendicular to $AI$, with $I$ the midpoint of $XY$. Now note that \[\angle{IRS}=\angle{CRS}=\frac{1}{2}\angle{ABC}=90^\circ - \frac{1}{2}\angle{BAC} - \frac{1}{2}\angle{ACB}=\angle{JCB}-\angle{RCB}=\angle{RCJ}=\angle{RSJ}.\] Hence $JS \| IR$ and, by the same argument, $JR \| IS$. Hence $JRIS$ is a parallelogram and if $K$ is the midpoint of $RS$, then $J$, $K$ and $I$ are collinear. It is well known that $R$ is the circumcenter of $\triangle{AIB}$ and that $S$ is the circumcenter of $\triangle{AIC}$. Hence $RS$ is the perpendicular bisector of $AI$. Now, since $AI \perp XY$ and $AI \perp RS$, it follows that $RS \| XY$. Since $I$ and $K$ are the midpoint of $XY$ and $RS$, respectively, it follows that $\mathcal{H}(I)=K$. Therefore, $K$, $I$ and $P$ are collinear, which implies that $P$, $I$ and $J$ are collinear. $\square$ Now let the midpoints of $BI$ and $CI$ be $U$ and $V$, respectively. Let the circumcircles of $\triangle{BUD}$ and $\triangle{CVD}$ intersect at $P' \neq D$. Note that $U$ and $V$ are the circumcenters of right triangles $\triangle{BID}$ and $\triangle{CID}$, respectively. Hence $\angle{UP'B}=\angle{UP'D}=\frac{1}{2}\angle{ABC}$ and $\angle{VP'D}=\angle{VP'C}=\frac{1}{2}\angle{ACB}$. Summing these angles yields that $\angle{BP'D}=180^\circ - \angle{BAC}$, which implies that $ABP'D$ is cyclic. Summing also yields that $\angle{UP'V}=90^\circ - \frac{1}{2}\angle{BAC}=180^\circ - \angle{BIC}=180^\circ - \angle{UBV}$, which implies that $IUP'V$ is cyclic. Since $U$ and $V$ are the midpoints of $BI$ and $CI$, it follows that $UV \| BC$ and hence that $\angle{IP'U}=\angle{IVU}=\frac{1}{2}\angle{ACB}$. Therefore $\angle{IP'B}=90^\circ - \angle{BAC}$, which implies that $J$, $I$ and $P'$ are collinear. By the lemma, since $P'$ lies on $\Omega$, this implies that $P'=P$. Now let $PD$ intersect $\Omega$ at $M'$. Since $PBUD$ is cyclic, $\angle{BPD}=\angle{ABC}$ which implies that the circumcircle of $PBUD$ is tangent to $AB$ and hence that $\angle{BDP}=180^\circ - \angle{PBA}=\angle{PM'A}$. Hence $M'A \| BD$ which implies that $M'=M$. Hence $P$, $D$ and $M$ are collinear. $\blacksquare$

darn this problem is "well-known" now