Oops, this was for p6.

Let $\angle CIQ = \varphi$ and $\angle BIP=\theta$. Clearly, we have that $\theta+ \varphi = 90^{\circ}-\frac{\alpha}{2}.$ Let $PB\cap(ABC)=T.$ Claim 1: $T\in CQ.$ Proof: It suffices to show that $\angle TCI = \angle IAQ.$ Now angle chasing $\angle TBI = \angle PAI = \theta+\frac{\alpha}{2}$ and since $\angle BTC = 180-\alpha,$ we get that $$\angle TCI = 360^{\circ}-\angle BTC - \angle TBI - \angle BIC = 90^{\circ}-\theta.$$However, $\angle IAQ = \frac{\alpha}{2}+\varphi = 90^{\circ}-\theta,$ so we are done. Claim 2: $T$ is the circumcenter of $\triangle APQ.$ Proof: Firstly, note that $\angle CQI = \angle CAI = \angle BAI = \angle BPI$. Hence, $\angle TPQ = \angle TQP$, i.e. $TP=TQ$. Moreover, we have that $\angle ATC = \angle ABC = \beta$ and since $\angle CQA = \frac{\alpha+\gamma}{2}=90-\frac{\beta}{2},$ we get that $\triangle TAQ$ is also isosceles. Thus, $TA=TQ=TP,$ finishing the problem.

Let $D=BP\cap CQ$. Then$$\measuredangle DPQ=\measuredangle BPI=\measuredangle BAI=\measuredangle IAC=\measuredangle IQC=\measuredangle PQD,$$so $DP=DQ$ and $\measuredangle PDQ=\measuredangle PAQ$, which means that $D\in (ABC)$. Then$$\measuredangle DPA=\measuredangle BPA=\measuredangle BIA=90^\circ -\measuredangle ICB$$and$$\measuredangle ADP=\measuredangle ADB=\measuredangle ACB=2\measuredangle ICB,$$so $\measuredangle PAD=90^\circ -\measuredangle ICB=\measuredangle DPA$. Hence $DA=DP=DQ$, so $D$ is the circumcentre of $APQ$, and it lies on the circumcircle of $ABC$. Done.

Cool problem. Let $M=CI\cap(ABC)$ and $N=BI\cap(ABC)$. It is well known that $A,B,C,M,N$ are concyclic. Also $M$ and $N$ are the circumcenters of $\triangle AIB$ and $\triangle AIC$ respectively. Now one needs to suspect that the circumcentre of $\triangle APQ$ can be found by the intersections of the lines $CQ$ and $PB$ and the rest is angle chasing. Let $S=PB\cap(ABC), \angle ABC=2b, \angle ACB=2c, \angle BAC=2a$. My method to show that $S,C,Q$ are collinear is the same as @rstenetbg(The first comment). Now we have $\angle BAI=\angle BPI=\angle CAI= \angle CQI=a$ which means $\angle SPQ=\angle SQP=a$ $\implies SQ=SP$ . Let $\angle CIQ=f$ then $\angle CIQ=\angle MIP=\angle MPI=f$ but since $\angle IAP=\angle IBA=b$ we will have $\angle MAP=\angle MPA=b-f$ and also $\angle MAB=\angle MBA=\angle MCB=c$. Now lets denote $\angle BAS=m$, since $ABSC$ is cyclic then $\angle SCB=m$. $\triangle SAP$ is isosceles $\iff$ $\angle SAP=\angle SPA \iff a+b=b-f+c+m \iff f+a=c+m$ which is true because $\angle MCS=\angle QIC+ \angle CIQ$ which is the same as $\angle SCB+ \angle MCB= a+f$ so $f+a=m+c$ therefore $SA=SP$ $\implies SA=SQ=SP$ $Q.E.D \square$

Attachments:

Solution during contest: Let $M$ be the midpoint of the arc $\overarc{AB}$ not containing $C$ and $O$ be the center of $(APQ)$. By the Incenter-Excenter Lemma we conclude that $M$ is the center of $(AIBP)$. Hence $M$ belongs to the perpendicular bissector of $AP$. $O$ also does because it is center of $(APQ)$. Thus $OM$ is the perpendicular bissector of $AP\rightarrow\angle{AOM}=\angle{POM}=\frac{\angle{AOP}}{2}$. Because $O$ is center, we also have $\angle{AQP}=\frac{\angle{AOP}}{2}\rightarrow \angle{AOM}=\angle{AQP}$. But $\angle{AQP}=\angle{AQI}=\angle{ACI}$, by the inscribed angle. Finally, $\angle{ACI}=\angle{ACM}$, because $C, I, M$ are collinear. Hence $\angle{AOM}=\angle{ACM}\rightarrow O\in(ACM)=(ABC)$, as we wanted.