$\textbf{LEMMA:-}$ $ABC$ be a triangle and $I$ be the Incenter of $\Delta ABC$. Let $\overline{BI}\cap\overline{AC}=\{E\}$ and $\overline{CI}\cap\overline{AB}=\{F\}$. Let $M$ be the midpoint of $BC$. Then the $A-$ Symmedain , $\overline{EF},\overline{IM}$ are concurrent. Proving this Lemma synthetically is not tough (See the second claim in (#3here Iran TST 2020 Day 1 P3), but I present a bary one. Let $A$- Symmedian $\cap\overline{IM}=\{X\}$. Let the Barcentric Coordinate of $X$ be $(x:y:z)$. Then clearly $y:z=b^2:c^2\implies z=\frac{c^2y}{b^2}$. Now as $X,I,M$ are collinear we get \begin{align*} 0=\begin{vmatrix} x & y & z \\ a & b & c \\ 0 & 1 &1 \\ \end{vmatrix} \implies 0=x(b-c)-ya+za=x(b-c)-ya+\left(\frac{ac^2y}{b^2}\right)\implies x(b-c)=y\left(a-\frac{ac^2}{b^2}\right)\implies x:y=\frac{a(b^2-c^2)}{b^2(b-c)}=\frac{a(b+c)}{b^2}\implies(x:y:z)=\left(ab+ac:b^2:c^2\right) \end{align*} So in order to prove $X,E,F$ collinear, we need the following determinant to be $0$. \begin{align*} \begin{vmatrix} a & 0 & c\\ a & b & 0\\ ab+ac & b^2 & c^2\\ \end{vmatrix}=abc^2+c(ab^2-ab^2-abc)=0\implies X,E,F\text{ are collinear} \end{align*}____________________________________________________________________________________________ Coming back to the problem. Notice that from our $\textbf{LEMMA}$ we have that $\overline{AQ}$ is the $A-$ Symmedian of $\Delta ABC$. Let $\{\overline{AP},\overline{AQ}\}\cap\odot(ABC)=\{T,V\}$ and $\overline{EF}\cap\odot(ABC)=\{X,Y\}$. Let $\overline{VT}\cap\overline{BC}=K$. So, $-1=(A,V;B,C)\overset{T}{=}(\overline{AT}\cap\overline{BC},K;B,C)\implies K\in\overline{XY}\}$. So, $\{(B,C),(X,Y),(T,V)\}$ are pairs of an Involution $\Phi$ and projecting $\Phi$ from $A$ on $\overline{EF}$ we get that $\{(X,Y),(E,F),(P,Q)\}$ are pairs of an Involution. Hence, $\odot(ABC),\odot(AEF),\odot(APQ)$ are Coaxial. $\blacksquare$

I finish this problem by barycentric coordinates. Since in barycentric coordinates, circle equation $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$ $u,v,w$ are the power to each vertice w.r.t the circumcircle. We can easily compute $(AEF)$ by bash length. For $(APQ)$, let $(APQ):-a^2yz-b^2zx-c^2xy+(x+y+z)(py+qz)=0$ and then substitute $P=(2a:b:c),Q=(ab+ac:b^2:c^2)$ into equation of $(APQ)$ and solve. Since the radical axis in bary are the subtraction of two circles, then $(AEF),(ABC),(APQ)$ are coaxial $\Leftrightarrow p:q=c^2(a+c):b^2(a+b)$. Then 10 minutes bash we can get the result.

Let $AI\cap BC=D$. Then projecting line $BC$ onto line $EF$ through $I$ we have $(BC;DM)=(EF;PQ)$. Hence $\frac{BD.CM}{CD.BM}=\frac{EP.FQ}{FP.EQ}$ ie $\frac{AB}{AC}=\frac{AE}{AF}.\frac{FQ}{QE}$. Hence $\frac{FQ}{QE}=\frac{AB}{AC}.\frac{AF}{AE}=\frac{FB}{EC}=\frac{RF}{RE}$ where the 2nd last equality holds true due to angle bisector thm applied twice and last one due to $R$ being centre of spiral similiarity taking $EC$ to $FB$. Hence by converse of angle bisector thm, $RQ$ bisects $\angle ERF$. Thus, if $L$ is midpoint of arc $EF$ of $\odot (AEF)$ not contaning $A$ then $L,Q$ and $R$ are collinear. Now by applying shooting lemma we have $LQ.LR=LP.LA$. Hence $APQR$ is cyclic as desired.

The conclusion still holds if we replace the incenter $I$ with any point on the bisector of $\angle BAC$.

Two line solution using ratio function: Define $f(X)=\frac{XF}{XE}$. We may prove $f(A)f(R)=f(P)f(Q)$ which suffices: $$f(A)f(R)=f(Q)\frac{BF}{EC}=f(Q)\frac{\frac{BC.FI}{IC}}{\frac{BC.IE}{BI}}=f(Q)\frac{IF}{IE}\frac{BI}{IC}=f(Q)\frac{IF}{IE}.\frac{sin(FIP)}{sin(EIP)}=f(Q)f(P)$$

Let $AI\cap BC=D$. Claim: $AQ$ is $A-$symmedian. Proof: Let $AM\cap EF=G,EF\cap BC=T$. Apply DDIT on quadrilateral $BCAEFI$ with center $M$ which implies $(\overline{MI},\overline{MA}),(\overline{ME},\overline{MF}),(\overline{MB},\overline{MC})$ is an involution. Project this involution onto $EF$ to get that $(T,T),(Q,G),(E,F)$ is an involution. Since the reflections of $AT,AE$ over $AI$ are $AT,AF$ we conclude that this involution is reflection over $AI$. Thus, $AM,AQ$ are isogonal lines on $\measuredangle CAB$ which yields the result.$\square$ Perform $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle CAB$. Note that $E^*C\parallel BI,BF^*\parallel CI$. By the claim, $A,M,Q^*$ are collinear. Let $(AEF)\cap BC=K,L$. We want to show that $KL,E^*F^*,P^*Q^*$ are collinear which is equavilent $(AK,AL),(AB,AC),(AI,AM)$ to be an involution since $K,L,E^*,F^*,P^*,Q^*$ are concyclic. Project this onto $BC$. We want to prove that $(D,M),(B,C),(K,L)$ is an involution. Let $N$ be the midpoint of arc $BAC$. $A,N,K,L$ are concyclic since $T,E,F$ are collinear. Also $N$ lies on the circle with diameter $ND$ which is $(ADM)$ and it lies on $(ABC)$. This yields $(ADM),(ABC),(AKL)$ are coaxial thus, $(D,M),(B,C),(K,L)$ is an involution as desired.$\blacksquare$