[asy][asy] size(9cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair P = dir(130); pair S = intersectionpoints(A--B,circumcircle(A,P,I))[1]; pair T = intersectionpoints(A--C,circumcircle(A,P,I))[1]; pair K = extension(P,S,I,I+dir(90)*(I-circumcenter(A,P,I))); pair L = extension(K,I,P,T); pair D = extension(B,K,C,L); draw(unitcircle); draw(A--B--C--A--cycle); draw(B--D--C); draw(P--K--L--P--cycle); draw(S--T); draw(circumcircle(A,P,I)); draw(B--I--C); draw(circumcircle(P,D,K),red+dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(90)); dot("$D$",D,dir(D)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$P$",P,dir(P)); dot("$K'$",K,dir(315)); dot("$L'$",L,dir(L)); [/asy][/asy] Let $S$ and $T$ be the intersection of $\omega$ with $AB$ and $AC$ respectively, and let $K'$ and $L'$ be the intersection of $PS$ and $PT$ with the tangent to $\omega$ at $I$. We prove that $BK'=K'I$ and $CL'=L'I$ which shows that $K\equiv K'$ and $L\equiv L'$. Throughout this proof, we use directed angles modulo $180^\circ$ to avoid configuration issues. Claim 1: $\triangle BSP\sim \triangle CTP$. Proof: \[\measuredangle BSP=\measuredangle ASP = \measuredangle ATP = \measuredangle CTP\]\[\measuredangle PBS=\measuredangle PBA=\measuredangle PCA=\measuredangle PTA \] Claim 2: $\triangle BK'P\sim \triangle CL'T$. Proof: We know that $\measuredangle BSK'=\measuredangle BSP=\measuredangle CTP=\measuredangle CTL'$, and they must be in fact equal in magnitude because $K'$ and $L'$ both lie on the same side of $ST$. Also, $SI=IT$ since $AI$ bisects $\angle SAT$, therefore $ST\parallel K'L'$. \[\frac{BS}{SK'}=\frac{BS}{SP}\cdot\frac{PS}{SK'}=\frac{CT}{TP}\cdot\frac{TP}{TL'}=\frac{CT}{TL'}.\] Let $BK'$ intersect $CL'$ at $D$. Note that since \[\measuredangle ABD=\measuredangle SBK'=\measuredangle TCL'=\measuredangle ACD,\]this means that $D$ lies on $(ABC)$. Claim 3: $D, P, K', L'$ are concyclic. Proof: \[\measuredangle PK'D=\measuredangle SK'B=\measuredangle TL'C=\measuredangle PL'D.\] Now, if I were to show $\triangle PAD\sim \triangle PSB$, then we would be done. The result would follow because \[\measuredangle K'BP=\measuredangle DBP=\measuredangle DAP=\measuredangle BSP\]Therefore, $K'B$ is tangent to $(BSP)$, and \[K'B^2=K'S\cdot K'P=K'I^2\]and hence $K'B=K'I$. Finally, taking a homothety from $P$ sending $ST$ to $K'L'$ also sends $\omega$ to $(PKL)$, finishing the problem.

Let $M_A$, $M_B$ and $M_C$ be the midpoints of arcs $BC,CA,AB$ (not containing the opposite vertex). Recall the fact that $M_BM_C$ is the perpendicular bisector of $AI$, etc. So we can rewrite the problem taking $M_A,M_B,M_C$ as the reference triangle: Quote: Let $\triangle ABC$ be a triangle with orthocenter $H$, and let $P$ be a point on its circumcircle. Let $R$ be the second intersection of $AH$ and $(ABC)$. Let the tangent to $(HPR)$ through $H$ meet $AB$ at $K$ and $AC$ at $L$. Prove that $(PLK)$ is tangent to $(PHR)$. Observe that \[ \angle PHK = \angle PRH = \angle PRA = 180 - \angle KBP, \]hence $PBKH$ is cyclic. Similarly, $HCLR$ is also cyclic. Now, we consider the (negative) inversion around $H$ sending $A,B,C$ to the feet of the altitudes. We can now rewrite the problem for a second time (this time i'll mantain the point names): Quote: Let $ABC$ be a triangle with incenter $H$, and let $AH$ meet $(ABC)$ again at $R$. Let $P$ be a point on $(ABC)$. If $L\in (AHC)$ and $K\in (AHB)$ such that $H\in LK$ and $RP\parallel LK$, prove that $PL = PK$. So now there's no weird circles to take care of. Notice that the cyclic quads we proved earlier now translate into $P,C,L$ being collinear and $P,B,K$ being collinear. Now the problem is obvious: if $\ell$ is the bisector of $\angle LPK$, then $\ell$ is perpendicular to $PR$ (as $R$ is the arc midpoint), hence $\ell \perp LK$, implying that $PLK$ is isoscelles in $P$.

Let $AI$ meet the circumcircle at $M$. By Pascal's theorem, $BK$ and $CL$ intersect at a point $S$ that lies on the circle $(ABC)$. Let $P'$ be the second intersection of circles $(SKL)$ and $(SBC)$. Let $P'D$ meet $BC$ at $D$ In the spiral similarity $\triangle P'KL \sim P'BC$, $$\frac{KI}{IL} = \frac{KB}{CL} = \frac{FB}{FC} = \frac{BD}{DC}$$So we actually have $P'KLI \sim P'BCD$. Next, $$\measuredangle P'AI= \measuredangle P'AM =\measuredangle P'DB =\measuredangle P'IK$$which means that $P'AI$ is tangent to $KL$ at $I$ and $P'=P$ Now, just note that $PI$ is the angle bisector of $\angle KPL$ due to similarity and by Archimede's lemma, $(PKL)$ is tangent to $(API)$

Let $M_b, M_c$ be midpoints of the minor arcs $AC, AB$. We have to prove that $PI$ bisects $\angle KPL$. We have that $\angle KM_cI=\angle KM_bI$, so we will be done if we prove that $IKM_cP$ and $ILM_bP$ are cyclic. Indeed, $\angle PIL=\angle PAI=\angle KM_cP$, done.