suffices to prove (P,B;M,AD\capBP)=-1 <=> (A,B;C,D)=-1, which is true.

Because $MB$ is tanget to circle $\Omega$ it is well know that: $MB^2 = MC\cdot MA$ but because $MB=MP$ we have: $MP^2 = MC\cdot MA$ Now this means that triangles $\triangle MPC$ and $\triangle MPA$ are similar. From there, we got: $ \angle MPC =\angle PAM$ But, line $PA$ is tanget to the circle $\Omega$ so: $\angle PAM= \angle ADP$. Now : $\angle MPC= \angle ADP$ so the lines $AD$ and $BP$ are parallel.

In this proof, let $(ABC$) be the circumcircle of $\triangle ABC$ and Pow$_\omega (X)$ be the power of point $X$ with respect to circle $\omega$. We start out with a Lemma. Lemma: Given $\triangle ABP$ with $M$ on the midpoint of $PB$ and $MB$ is tangent to $(ACB)$, then $MP$ is tangent to $(ACP)$. Proof of Lemma: Let $\Omega=(ACB)$ and $\omega=(ACP)$. Because $M$ lies on the Radical Axis $AC$ of $\Omega$ and $\omega$, Pow$_\Omega (M)=$Pow$_\omega (M).$ Assume for the sake of contradiction that $MP$ is not tangent to $\omega$, or $MP$ intersects $\omega$ at a second time at $P'$. Because $MP^2=MB^2$ and Pow$_\Omega (M)=MB^2$, then Pow$_\omega (M)=MP^2$. But this is a contradiction as Pow$_\omega (M)=MP\cdot MP' \neq MP\cdot MP=MP^2$ as $P$ and $P'$ are unique. Thus, $MP$ is tangent to $(ACP).$ $\Box$ By the Lemma, $\angle BPC=\angle PAC.$ Because $PA$ is tangent to $\Omega$, $\angle PAC=\angle CDA$. Hence, $\angle ADC=\angle PAC=\angle CPM$, proving that $AB\parallel PB$. $\blacksquare$

Since, $\Delta MPC \sim \Delta MAP$ and $\Delta PAC \sim PDA \implies \angle CPM =\angle PAM= \angle PDA \implies \boxed{AD||BP}$ Something obvious but interesting is that: $PC=2MC$

Define $\mathcal{P_1}_{||AD}$ as the point at infinity along $AD$, $$-1 =(A,B;C,D) \overset{A}{=} (P,B;M, AD \cap BP) \implies AD \cap BP=\mathcal{P_1}_{||AD} \implies AD||BP$$

Here's a power of a point solution. Claim: $\triangle MPC \sim \triangle MAP$ Proof: $PM^2 = BM^2 = CM \cdot AM,$ so $\frac{MP}{MC} = \frac{MA}{MP}.$ Notice that because $PM = \frac{AP}{2}, PC = 2CM.$ Now $PC \cdot PD = PA^2 = PB^2 = 4MB^2 = 4(MC \cdot MA)$. But $PC = 2MC,$ so $PD = 2AM.$ Now $\frac{AC}{CD} = \frac{AM-CM}{PD-CD} = \frac{1}{2},$ so $\triangle ACD \sim \triangle MCP,$ and so $\angle MPC = \angle CDA = \angle PDA,$ as desired.

Since $PB$ is tangent to $\Omega$ and $AC$ intersects $BP$ in its midpoint we have that $(PCA)$ is tangent with $PB$ and so, $\angle CDA= \angle CAP=\angle BPC=\angle BPD$ hence done.

Solution. It is easy to see that $(ACP)$ is tangent with $PB$ at point $P\implies C$ is the $A$-Humpty Point of $\triangle ABP$, $$ \angle PCB=180-\angle A\implies \angle PBA=\angle A=\angle DCP=\angle DAB.\blacksquare$$

first by pow(M) MB^2=MC*MA=CP^2 because M is midpoint.so BP is tangent to (ACP) which means (1) <CAP=<CPB.PA is tangent to circle so <CAP=<ADC <=> (1) <ADP=<BPD and we are done.

$\color{red} \textbf{Geo Marabot Solve 4}$ $$-1=(D,C ; AB \cap PD,P) \overset{A}{=} (AD\cap BP,M;B,P) \implies AD\cap BP \equiv \mathcal{P_1} \implies AD \parallel BP \blacksquare$$