Here's the full problems and solution link: http://igo-official.ir/wp-content/uploads/2016/12/igo_2016_-_problems__solutions_-_english.pdf

We will use the well-known fact that the tangent to the circumcircle of the triangle $ABC$ divides the side $BC$ in ratio $-\frac {AB^2}{AC^2}$. Before applying it, let note a Lemma. Lemma:$\frac{AQ}{CQ}=\frac{AB}{AC}$. Proof: This can be proven considering the inversion with center $P$ and radius $PA$, and as the circle of inversion is orthogonal to $\omega$, $(M, Q)$ and $(B,C)$ are excanging places. So, by applying the inverison distance formula, we get $\frac{AQ}{CQ}=\frac{AM\times BP}{BM\times AP}=\frac{BP}{AP}=\frac{AB}{AC}$. Further, we easily compute the ratios $\frac{AK}{KC}=\frac{AQ^2}{CQ^2}=\frac{AB^2}{AC^2}=\frac{BP}{CP}$, and by Thales, we have the conclusion.

I too have inversion solution but it is different. Let $K'$ be on $AC$ such that $\angle PK'C = 90^{\circ}$. Let $N$ be midpoint of $BC$. We shall prove that $K'Q$ is tangent to $\omega$. Perform an inversion with centre $P$ and radius $PA$. Then $\omega$ gets mapped to $\omega$ itself. $K'$ goes to $PK' \cap AN$ say $L$ and $Q$ goes to $M$. So line $K'Q$ gets mapped to $\odot (PLM)$. Now note that $MN$ is perpendicular bisector of $AB$ and hence $MN$ is perpendicular bisector of $PL$. However circumcentre of $\odot (PLM)$ lies on perpendicular bisector of $PL$. Also $N$ is circumcentre of $\omega$. $\therefore$ The centres of $\omega , \odot (PLM)$ and their intersection pts. are collinear. So $\omega$ and $\odot (PLM)$ are tangent to each other. $\therefore$ before inversion $K'Q$ and $\omega$ must have been tangent to each other as desired.

I am not sure if this works for $AB>AC$ but I hope it does. Let $AQ$ and $MC$ meet at $R$. Pascal on $AQQMCA$ shows that $R$, $K$ and $P$ are collinear. Pascal on $ABCMMQ$ shows that $PR$ is parallel to $AB$. Thus $PK$ is parallel to $AB$ and $\angle PKC=90^{\circ}$.

@above, how can you use pascal if the lines $AB$ and $MM$ do not even intersect? I have not learned about this use of pascal before, so could you explain?

we have to prove PK || AB or PB/PC = KA/KC. first note that PMA and PAQ are similar. PMB and PCQ are similar. PBA and PAC are similar. PB/PC = (AB/AC)^2 and KA/KC = (QA/QC)^2 so we have to prove AB/AC = QA/QC. AB/AC = PB/PA PA/PQ = AM/AQ and PQ/PB = QC/BM so PB/PA = QA/QC. we're Done.

Euler365 wrote: I too have inversion solution but it is different. Thank you for your very good inversion solution!!

Finally Solved With $\textbf{Pascal's Theorem}$ $\color{red} \boxed{\textbf{SOLUTION}}$ Let, $AQ \cap MC \equiv X, AB \cap AC \equiv Y$ By $\textbf{Pascal's Theorem}$ on $Q,Q,A,A,C,M$ We get $QQ \cap AC \equiv K, QA \cap CM \equiv X, AA \cap MQ \equiv P$ So, $P,X,K$ are collinear. We need to show $PK \equiv PX \parallel AB$ As $M$ is the midpoint of minor arc $AB, CM$ is the angle bisector of $\angle ACB$ Let, $\angle ACM=\angle BCM=\alpha \implies \angle ABC=90-2\alpha$ Now, $$\angle PQX=\angle AQM=\angle ACM=\angle BCM=\angle PCX$$$\implies PQCX$ cyclic Therefore, $$\angle CXK=\angle PQC=\angle PQX + \angle XQC = \alpha + \angle ABC=\alpha + 90- 2\alpha=90-\alpha$$So, $$\angle AYC=90- \alpha= \angle CXK=\angle YXK \implies AY \parallel PX \implies AB \parallel PX \implies AB \parallel PK \implies \angle PKC=\angle BAC=90 \blacksquare$$

It suffices to show that $PK\parallel AB$. This can be done with 3 applications of Pascal's Theorem. Let $T=MA\cap QB$ and $X=QA\cap CM$. Pascal's on $QQMACB$ yields $K, P, T$ are collinear. Pascal's on $ABCMMQ$ yields $PX\parallel AB$ (since $MM$ intersects $AB$ at the point at infinity along that line) Pascal's on $QQAACM$ yields $K,X,P$ are collinear. Therefore, $K,P,T,X$ are collinear and these lie on a line parallel to $AB$, so we are done.

It's just angle chasing