Let $N$ be the midpoint of arc $BAC$ and let $Q'$ be the intersection of $AN$ and $TM$. It is a well-known fact that $T, I$ and $N$ are collinear, and therefore by Pascal's theorem on $AAMTTN$ the points $Q' = AN \cap TM, I = AM \cap TN$ and $P = AA \cap TT$ are collinear. Therefore $Q = Q'$ and the result follows.

The problem becomes a Trivial orthocenter configuration lemma WRT $\Delta Q'MM'$, where $TM \cap AM'=Q'$ and $M'$ is midpoint of arc $BAC$

Let the external bisector of $\angle A$ intersect $\Gamma$ at $N{}$. It is well-known that $N,I$ and $T{}$ are collinear. Further, as $N{}$ is the midpoint of the arc $BAC$ then $M,O$ and $N{}$ are collinear. Hence, we want to show that $Q,A$ and $N{}$ are collinear, that is, we want to show that $AQ$ is the external bisector of $\angle A$. Therefore, it suffices to show that $QA$ is perpendicular on $AI$. Claim: The line $PI$ is parallel to $BC$. Proof: Perform the inversion centered at $P{}$ which leaves $\Gamma$ unchanged. The claim then becomes equivalent to this. $\square$ Let $PI$ intersect $\Gamma$ at $X{}$ and $Y{}$. As $AI$ is the bisector of $\angle BAC$ and $XY$ is parallel to $BC$, then $AI$ is the bisector of $\angle XAY$ as well. Now, note that since $PA$ and $PT$ are tangent to $\Gamma$ then $AXTY$ is a harmonic quadrilateral. Projecting onto $PI$ through $M{}$ it follows that \[(X,Y;I,Q)=-1,\]but since $AI$ is the internal bisector of $\angle XAY$ then $AQ$ is the external bisector of $\angle XAY$ so $AQ$ is perpendicular on $AI$ as desired.