It seems like Anant is reviving his configuration... Claim 1: $Q\in (BFK),(CFK)$. By some simple angle chasing we can see that $$\angle KEA =\angle KDE =90^{\circ}-\angle FED =\frac{\angle B}{2}=\angle IBC =\angle IQC,$$so $KQCE$ is cyclic, and so must $KQBF$ be. $\square$ Claim 2: $\overline{B,Q,S},\overline{C,Q,R}$ are collinear. As above, $\angle KQC =\angle IBC=\frac{\angle B}{2}$ and $\angle KQR =\angle KFE=\angle KQC =\frac{\angle B}{2}$, so $\overline{C,Q,R}$ must be collinear. The same applies for $\overline{B,Q,S}$. $\square$ Claim 3: $BICP$ is harmonic. As noted in IMO 2019 P6, $L$ is the point of tangency between the $A$-mixtilinear incircle and $\Gamma$. By a well-known lemma, $LI$ passes through the top-most point of $\Gamma$. By Incircle-Excircle Lemma, the tangents in $B$ and $C$ to $(BIC)$ must also pass through this point, proving that $BICP$ is indeed harmonic. $\square$ Claim 4: $P$ is the Miquel Point of $BCRS$. To see this, it's enough to prove that $P$ is the center of spiral similarity mapping $SB$ to $RC$, since $P$ is already on $(BCQ)$. But by the Law of Sines in $\triangle BFS$ and $\triangle CER$ we have $$\frac{BS}{CR}=\frac{BS/\sin(\angle AFE)}{CR/\sin(\angle AEF)}=\frac{BF/\sin(\angle ESQ)}{CE/\sin(\angle FRQ)}.$$However by some simple angle chasing, $\angle ESQ =\angle EKQ =\frac{180^{\circ}-\angle KIE}{2}=90^{\circ}-\angle KDE=90^{\circ}-\frac{\angle B}{2}$, so continuing the ratios we have $$\frac{BS}{CR}=\frac{BF/\cos(\angle B /2)}{CE/\cos(\angle C/2)}=\frac{BI}{CI}=\frac{BP}{CP}.$$So we have proved that $\frac{SB}{CR}=\frac{BP}{CP}$. Since $\angle SBP=\angle RCP$ (because $BPQC$ is cyclic), $\triangle PBS\sim\triangle PCR$, exactly what we wanted to prove. $\square$ Since $P$ is the described Miquel Point, it must lie on $(QSR)$, finishing the proof. $\blacksquare$

Its a nice one Let $X$ the point of tangency of the A-excircle with $BC$ and now note that since by homothety $AX,ID,\omega$ are concurrent then $AX,AK$ are isogonal but by by $\sqrt{bc}$ inversion u see that $L$ has to be the A-mixtilinear intouch point with $\Gamma$, let $N$ the midpoint of arc $BAC$ in $\Gamma$ then clewrly $NB,NC$ are tangent to $(BIC)$ and by $\sqrt{bc}$ inversion $N,I,L$ are colinear so $BICP$ is harmonic. Now let $S_A$ the A-sharkydevil point, note that by the spiral similarity and radax $S_AI,KD,EF$ are concurrent by by spiral sim u also see that $S,D,M$ are colinear where $M$ the midpoint of minor arc $BC$ in $\Gamma$ hence $K$ is mapped to $I$ on the spiral similarity that sends $EF$ to $BC$ and now by angle chase $\angle AEK=\angle NCI=\angle IBC=\angle KQC$ so $Q$ lies in $(KEC)$ and in a similar way u see that $Q$ lies in $(KFB)$ and now by more angle chase $\angle SQC=\angle AEF=\angle AFE=\angle BQC$ so $S,B,Q$ are colinear and similarily $R,C,Q$ are colinear now by even more angle chasing (this time we need some remainders on ortocenter config over $\triangle DEF$ lol) $\angle FSB=\angle EKI=\angle FED=90-\angle IBC$ and in a similar way $\angle ERC=90-\angle ICB$ and now by ratio chase. $$\frac{SB}{CR}=\frac{FB}{EC} \cdot \frac{\cos(\angle BAI) \cdot \cos(\angle ICB)}{\cos(CAI) \cdot \cos(\angle IBC)}=\frac{BD \cdot \cos(\angle ICB)}{DC \cdot \cos(\angle IBC)}=\frac{BI}{IC}=\frac{BP}{PC}$$And since $BPQC$ is cyclic we get that there is a spiral similarity that sends $RS$ to $BC$ centered at $P$ which gives $PQRS$ cyclic as desired, thus we are done

solved with crazyeyemoody907 geometry is keeping us awake also shoutout to Pranav1056 because this is my 1056th post lmao [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=mediumblue; pen sec=purple; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=paleblue; pair O,A,B,C,I,D,E,F,U,K,L,P,Q,R,S,N; O=(0,0); A=dir(110); B=dir(210); C=dir(330); I=incenter(A,B,C); D=foot(I,B,C); E=foot(I,A,C); F=foot(I,A,B); U=foot(D,E,F); K=intersectionpoint(D+0.1*(U-D)--D+100*(U-D),incircle(A,B,C)); L=intersectionpoint(A+0.1*(K-A)--A+100*(K-A),circumcircle(A,B,C)); P=intersectionpoint(I+0.1*(L-I)--I+100*(L-I),circumcircle(B,I,C)); Q=intersectionpoint(K+1.1*(I-K)--K+100*(I-K),circumcircle(B,I,C)); S=extension(Q,B,E,F); R=extension(Q,C,E,F); N=intersectionpoint(L+0.1*(I-L)--L+100*(I-L),circumcircle(A,B,C)); draw(A--B--C--cycle,pri); draw(circumcircle(A,B,C),pri); draw(incircle(A,B,C),pri); draw(D--E,pri); draw(D--F,pri); draw(E--K,pri); draw(F--K,pri); draw(I--P,pri); draw(circumcircle(B,I,C),sec); draw(K--Q,pri); draw(R--S,pri); draw(Q--B,dashed+pri); draw(Q--C,dashed+pri); draw(B--S,dashed+tri); draw(C--R,dashed+tri); draw(circumcircle(P,Q,R),dashed+tri); draw(I--N,pri); draw(I--B,dashed+tri); draw(I--C,dashed+tri); draw(P--B,sec); draw(P--C,sec); label("$A$",A,dir(110)); label("$B$",B,dir(210)); label("$C$",C,dir(330)); label("$I$",I,dir(150)); label("$D$",D,dir(270)); label("$E$",E,dir(70)); label("$F$",F,dir(120)); label("$K$",K,dir(90)); label("$L$",L,dir(300)); label("$P$",P,dir(240)); label("$Q$",Q,dir(300)); label("$R$",R,dir(60)); label("$S$",S,dir(150)); label("$N$",N,dir(90)); [/asy][/asy] First, by angle chasing $\triangle{KFE} \sim \triangle{IBC}$, and if $N$ is the midpoint of $\widehat{BAC}$ then $AFEK \sim NBCI \implies$ the angle between $AK, NI$ = the angle between $EF, BC \implies N, I, L$ are collinear, and $PBIC$ is harmonic. Also, $\angle{SQC} = \angle{FEA} = \angle{BQC} \implies S, B, Q$ are collinear and so are $R, C, Q$. Thus, by spiral similarity from $(QBC)$ to $(QRS)$, it suffices to show that $\triangle{PBS} \sim \triangle{PCR}$, or $\frac{PB}{SB} = \frac{PC}{CR}$. $\frac{PB}{PC} = \frac{IB}{IC}$, so it suffices to show that $\frac{IB}{BS} = \frac{IC}{CR}$, which is obvious with a little length chasing since $\triangle{DEF} \sim \triangle{FSB} \sim \triangle{ECR}$. $\square$

Note that $L$ is $A-$mixtilinear touch point hence if $N$ is the midpoint of arc $BAC$, then $N,I,P$ are collinear. Let $EF\cap BC=G$. Claim: $Q\in (KFB),(KEC)$. Proof: \[\measuredangle KQC=\measuredangle IQC=\measuredangle IBC=\frac{\measuredangle B}{2}=\measuredangle KDE=\measuredangle KEA\]Thus, $K,E,C,Q$ are concyclic. Similarily we conlude that $K,F,B,Q$ are concyclic.$\square$ Claim: $Q,B,S$ and $Q,C,R$ are collinear. Proof: Since $K,Q,C,E,S$ are concyclic, we have \[\measuredangle SQC=\measuredangle SEA=90-\frac{\measuredangle A}{2}=\measuredangle BQC\]So $S,B,Q$ are collienar and the other follows similarily.$\square$ Claim: $P,B,S,G$ are concyclic. Proof: Set $PG\cap (BIC)=H$. We observe $(G,D;B,C)=-1=(P,I;B,C)\overset{H}{=}(G,HI\cap BC;B,C)$ implies $H,D,I$ are collinear. Thus, \[\measuredangle GPB=\measuredangle HIB=\measuredangle DIB=90-\frac{\measuredangle B}{2}=\measuredangle EKQ=\measuredangle ESQ=180-\measuredangle BSG\]Which gives the result.$\square$ Since $P$ lies on both $(QBC)$ and $(GBS)$, we see that $P$ is the miquel point of $BCRS$ hence $P$ lies on $(QRS)$ as desired.$\blacksquare$