This point $K$ of intersection is point on $AA$ such a $KI \parallel BC$. The idea of my solution is to show that $K, M, P$ are collinear by two projections: on line $AA$ and $BC$ and then we know all lengths of segments and we can bash... I used the follow lemma: Let $l_1, l_2$ be lines and let $X, Y, Z$ be points such a numbers $Pr_X(l_i)Pr_Y(l_i)/Pr_X(l_i)Pr_Z(l_i)$ are equals for $i=1, 2$, there $Pr_T(l)$ is projection of point $T$ onto line $l$. Then $X, Y, Z$ are collinear. P.S. isn't it too hard for P11? It seems that in previous year it was easier...

Claim 1: $P,Q,B,C$ are cyclic Let $D$ be midpoint of $AL$. $J$ is A-excircle of $\triangle ABC$ and $A'=AJ \cap (ABC)$ From $$\measuredangle{QDJ}=\measuredangle{QCJ}=90=\measuredangle{JDP}=\measuredangle{JBP}$$ we get $Q,D,C,J$ and $P,D,B,J$ are cyclic. Now using power of point $$QI\cdot IC=DI\cdot IJ=PI\cdot IB$$We Prove our result. Define $B'=BI \cap (ABC),C'=CI\cap(ABC)$ Let $X=PQ \cap BC$ and $R=B'C' \cap AX$ Frist note as $XA=XL$ (As it's lie on perpendicular bisector) which give us $X$ is center of $A$-apollonius circle and $AX$ is tangent to $(ABC)$ by property of apollonius circle. by power of point $$XA^2=XB \cdot XC=XP \cdot XQ$$which mean $AX$ is also tangent to $(APQ)$ Claim 2: $I$ also lie on $(APQ)$ It well know that $(A,L;I,J)=-1$, now by harmonic property we get $$IA\cdot IL=DI\cdot IJ=PI\cdot IB=QI \cdot IC$$hence $A,Q,L,C$ and $A,P,L,B$ are cyclic. $$\measuredangle{QAP}=\measuredangle{QAL}+\measuredangle{LAP}=\measuredangle{QCL}+\measuredangle{LBP}=\measuredangle{CIB}=\measuredangle{QIP}$$ as $R$ lie on perpendicular bisector of $AI \Rightarrow RA=RI$ and from $RA$ tangent to $(APQ)$ we get $RI$ also tangent to $(APQ)$. 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dot((16.01234531245498,-10.402249410841005),linewidth(4.pt) + dotstyle); label("$K$", (16.16137444583791,-10.087708984927923), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let tangent at $P,Q$ to $(APQ)$ meet at $K$. Define $P'=RP \cap (APQ)$ and $Q'=RQ \cap (APQ)$ hence by harmonic property $(A,I;P',P)=(B,I;Q',Q)=-1$ From $\measuredangle{IPK}=\measuredangle{IQP}=\measuredangle{IAP}=\measuredangle{LBP}=\measuredangle{PBA}$ we get $PK \parallel AB$ $$(A,B;M,\infty{\overline{AB}})\stackrel{P}{=}(A,I;PM\cap(APQ),P)=-1$$ hence we get $P'=PM\cap(APQ)$ and hence $R,M,P$ are collinear and doing same for $Q$ we get $Q'=QN \cap (APQ)$ and hence $Q,N,R$ collinear. $\blacksquare$

Coordinate bash(mostly)

Let tangent to the circumcircle of $ABC$ at $A \cap BC = D$. Well known that $D$ lie on perpendicular bisector to the bisectrix $AL$ so $D, P, Q$ lie on one line bisectrix $\angle ADC$. Note that $P$ - incircle center $\Delta ABD$ and $Q$ - $D$-excircle center $\Delta ACD$. Let $I$ - incircle center $\Delta ABC$; $B_1, C_1$ - intersect $BI, CI$ with $(ABC)$; $\Omega_1, \Omega_2$ - circles with centers $B_1, C_1$ and tangents to $AC, AB$(at $N, M$). Then i need lemma(All-Russian MO 1999, 9-3): tangent to the circumcircle of $ABC$ at $A$ and $l$(line trought $I \parallel BC$) are common tangent to $\Omega_1, \Omega_2; S, T$ - tangent point $\Omega_1$ with $AD, l$; $K, L$ - tangent point $\Omega_2$ with $AD, l$; $X = AD \cap l$. Then note that $KM \parallel AP$(both $\perp AC_1$); $KL \parallel ST$(both $\perp B_1C_1$); $LM \parallel IP$ so $\Delta KLM \sim \Delta AIP$ and $M, P, X$ lie on one line. Similary $N, Q, X$ lie on one line. $\square$