Trivially from I-E Lemma it's enough to prove that $PQ$ is the perpendicular bisector of $II_A$, let $M$ midpoint of minor arc $BC$ and let $NI_A \cap EF=U$, first note that: \[\angle UTB=90-\frac{\angle A}{2}=\angle AFE=\angle AEF=\angle UTC \implies TBFU, TCEU \; \text{cyclic}\]Now we also have $\angle UIM=90=\angle UTM$ so $IUTM$ is cyclic, so by PoP: \[I_AB \cdot I_AP=I_AU \cdot I_AT=I_AC \cdot I_AQ=I_AI \cdot I_AM\]Which imply that $BPMI, CQMI$ are both cyclic and also since $M$ is center of $(BICI_A)$ we notice that this means $\angle IMP=90=\angle IMQ$ therefore not only $P,M,Q$ are colinear, in fact they lie on the perpendicular bisector of $II_A$, thus we are done .

Is P3 posted? Or is it from ISL?

Let $S = NI_a \cap EF$. From easy angle chase, we have cyclic pentagons $(ESTQC)$ and $(FSTPB)$. So we are done if we prove image of $A'$ in $\sqrt{I_a T \cdot I_a S}$ inversion lies on $(BIC)$. Let $I_aA' \cap (BIC)$ meet again in point $X$ and $M$ the midpoint of $II_a$. From Reim's, $SIMT$ cyclic, and also $MIA'X$ cyclic by angle chase. So the $I_aT \cdot I_aS = I_aA' \cdot I_a X$, done.

I thought I was smart when I noted that $T$ was the $A-$mixtillinear extouch point, but it actually turns out that knowing that doesn't help you much with the solution. Pretty straightforward for Israel standards. We denote by $X$ the intersection of lines $I_AN$ and $EF$. We start off with some easy observations. Claim : Point $X$ lies on $(BFT)$ and $(CET)$. Proof : Note that, \[\measuredangle BTX = \measuredangle BTN = \measuredangle BCN = \measuredangle AFE = \measuredangle AFX \]Thus, $BFXT$ must be cyclic. Similarly, $CEXT$ is also cyclic, and we have our claim. Now note that since lines $\overline{BP}$ , $\overline{CQ}$ and $\overline{XT}$ are concurrent at $I_A$, by the converse of the Radical Center Theorem, it follows that $BCQP$ is cyclic. Next, we can observe the following. Claim : Lines $\overline{EF}$ and $\overline{PQ}$ are parallel. Proof : Note that, \begin{align*} \measuredangle (BI_A,EF) &= \measuredangle EFA + \measuredangle I_ABC \\ &= \frac{\pi}{2} + \measuredangle IBC + \frac{\pi}{2} + \measuredangle CAI\\ &= \frac{\pi}{2} + \measuredangle ICB \\ &= \measuredangle I_ACB\\ &= \measuredangle QPI_A \end{align*}from which the claim follows. Next, we have the following rather convenient cyclic quadrilaterals. Claim : Quadrilaterals $BPMI$ and $CQMI$ are cyclic. Proof : Since $\measuredangle NAI = \frac{\pi}{2}$ it is clear that $AN \parallel EF$. Thus, $\triangle I_AIX \sim \triangle I_AAN$. Now, \begin{align*} \frac{I_AI}{I_AA} &= \frac{I_AX}{I_AN}\\ \frac{I_AI \cdot I_AM}{I_AA \cdot I_AM}&= \frac{I_AX \cdot IA_T}{I_AN \cdot IA_T}\\ I_AI \cdot I_AM &= I_AX \cdot I_AT\\ I_AI \cdot I_AM &= I_AB \cdot I_AP \end{align*}which implies that $BPMI$ is cyclic. Similarly it also follows that $CQMI$ is cyclic, proving the claim. Thus, $\measuredangle PMI = \measuredangle PBI = \frac{\pi}{2}$ and $\measuredangle IMQ = \measuredangle ICQ = \frac{\pi}{2}$ so points $P$ , $M$ and $Q$ are collinear. Now, let $A'$ be the $A-$antipodal point in $\Omega$. Since $\measuredangle AMA' = \frac{\pi}{2}$, it follows that $A'$ also lies on $\overline{PQ}$, and we are done.

BR1F1SZ wrote: Is P3 posted? Or is it from ISL? P3 was RMMSL.