I think it should be the tangent line at $C$ to the circumcircle of $ABC$? The tangent line from $C$ to $\omega$ just doesn't make sense at all since it's just the lines $CA$ and $CB$.

@above yes the tanget at $C$ should be to the circumcircle of $ABC$. Let $N$ be the midpoint of arc $ACB$. We will show that $N$ and $M$ lie on the $I$-symedian of triangle $AIB$. Firstly note that $\angle NAB = 90^{\circ} - \frac{C}{2} = \frac{A}{2} + \frac{B}{2}$ so $\angle NAI = \angle NAB - \angle IAB = \frac{B}{2} = \angle IBA$ so $NA$ is tangent to $(AIB)$. Similarly we get that $NB$ is tangent to $(AIB)$ so $N$ lies on the $I$-symedian of triangle $AIB$. Now we have that $\angle CDA = A - B$ so $\angle QDA = \frac{A-B}{2}$. Hence $\angle DQA = 180^{\circ} - \left(\frac{A-B}{2}\right) - \frac{A}{2} - (C+B) = \frac{B}{2} = \angle PBA$ so $PQBA$ is cyclic and so $\triangle{PIQ} \sim \triangle{AIB}$. Now let $T$ be the intersection of the $I$-symedian with $AB$. From the similarity we get that $\angle AIT = \angle MIQ$ so $T$, $I$, $M$ are collinear. Done!

ARMO 2015 11.7 https://artofproblemsolving.com/community/c6h1172751p5641839

Claim. $PQ\cap BC=\{X\}\Rightarrow I,B,X,Q$ are concyclic. Proof. By a trivial angle chase we get $\angle{ADP}=\frac{A-B}{2}$ and $\angle{PAD}=180-\frac{A}{2}$ which implies $\angle{IPQ}=\angle{APD}=\frac{B}{2}=\angle{IBX}\Rightarrow I,B,X,Q$ are concyclic. $\blacksquare$ Claim. Name the midpoint of arc $ACB$ as $N$. $CN\parallel PQ$. Proof. $\angle{PXC}=\angle{BIP}=180-\angle{AIB}=90-\frac{C}{2}$ from the first claim. $CN$ is the exterior bisector of angle $C$ so $\angle{XCN}=90-\frac{C}{2}=\angle{PXC}$ holds. This is enough to imply the claim. $\blacksquare$ The exterior bisector of angle $C$ is also $I_AI_B$ where $I_A$ and $I_B$ are the excenters of triangle $ABC$. $N$ is the midpoint of the line segment $I_AI_B$. By homothety we get that $IN$ bisects $I_AI_B\Leftrightarrow IN$ bisects $PQ$. This finishes our proof. $\blacksquare$