Let $K,L$ be the midarcs $AB,AC$ of the circumcircle $ABC$. Clearly $KL\perp AI$ and since $AK=KI$, $AL=LI$ we have $KL$ bisects $AI$ and thus bisects $IQ$ at $T$. Now Butterfly theorem gives $IT=IP$ applied to $ABC$ and $IO$ $\blacksquare$

For the sake of the times when complex bash was not so popular, let's do a smart trig bash! Without loss of generality let $AB < AC$. (Note that $Q$ is at infinity if $AB = AC$.) Denote $\angle AIO = \angle AQI = \varphi$. Then $QI = \frac{AI}{\sin \varphi}$ from the right triangle $AQI$ and $\sin\varphi = \frac{AO}{OI}\sin\frac{\beta-\gamma}{2} = - \frac{AO}{OI}\cos(\frac{\alpha}{2}+\beta)$ from the Sine Law in triangle $AIO$. To compute $IP$, apply the Sine Law on $IPC$ to get $IP = \frac{CI\sin\frac{\gamma}{2}}{\cos(\frac{\alpha}{2} + \beta - \varphi)}$. As $\frac{AI}{CI} = \frac{\sin\frac{\gamma}{2}}{\sin\frac{\alpha}{2}}$, the equality $IQ = 2IP$ reduces to $\cos(\frac{\alpha}{2} + \beta - \varphi) = 2\sin\varphi \sin\frac{\alpha}{2}$ and hence to $$ \cot \varphi = \frac{2\sin \frac{\alpha}{2} - \sin(\frac{\alpha}{2} + \beta)}{\cos(\frac{\alpha}{2}+\beta)}, \mbox{ where} \sin\varphi = -\frac{AO}{OI}\cos\left(\frac{\alpha}{2}+\beta\right). $$Now since $\varphi < 90^{\circ}$, using Euler's formula $OI^2 = R^2 - 2Rr$, we have $\cot\varphi = \sqrt{\frac{1}{\sin^{2}\varphi}-1} = \sqrt{\frac{OI^2/R^2}{\cos^2\left(\frac{\alpha}{2}+\beta\right)}-1} = \frac{\sqrt{\sin^2\left(\frac{\alpha}{2}+\beta\right) - \frac{2R}{r}}}{-\cos\left(\frac{\alpha}{2}+\beta\right)}$, whence it remains to show $$ \sin\left(\frac{\alpha}{2}+\beta\right) - 2\sin\frac{\alpha}{2} = \sqrt{\sin^2\left(\frac{\alpha}{2}+\beta\right) - \frac{2R}{r}}. $$After squaring and cancelling, we reduce to $\frac{R}{r} = 2\sin\frac{\alpha}{2}(\sin(\frac{\alpha}{2}+\beta) - \sin\frac{\alpha}{2})$ and hence to $r = 4R\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$, which is well known and we are done.

Let $BI$ and $CI$ meet circle $ABC$ at $K,S$. Note that $SK || AQ$ and $SK$ bisects $AI$ so $SK$ bisects $IQ$ at $T$. Note that we need to prove $IT = IP$. Let $PT$ meet circle of $ABC$ at $X,Y$. Note that we need to prove $XP = TY$ or $BP.PC = KT.TS$ or $\frac{BP}{KT} = \frac{PC}{TS}$ which is true since $I$ is intersection of $SC,KB$ and $O$ is its circumcenter.