Okay so this thing took me like 10 minutes to draw lol. Wording kinda threw me off After some angle chasing; this reduces to: show that $\angle LKP =45 $ deg We let the intersection point of lines SL and KP be X; and that of KL and SP be Y. This transforms the problem into showing that quadrilateral $SKXY$ is cyclic. Anyone can pick up?

Let $PK \cap AB=T$ and let $L'$ be the foot of tangent from $T$ to $w_1$ such that $L',P$ lie on the same side of $AB$.Obviously,$O,S,L',K,T$ are on a circle with diameter $OT$,Also,by POP: $$PK.PT=PS.PO \implies PO=PT$$Again,by POP for $T$: $$TA.TB=TK.TP=OP.OS=OA^2 \implies TL'=OA=OL' \implies \angle TOL'=\angle L'TO=45=\angle L'SP$$and since $L'$ is on $w_1$,we deduce that $L' \equiv L$.This ends the problem because $L$ is the midpoint of arc $SK$ in $(OSLKT)$ so $LK=LS$.

Let $C = OK\cap AB$, draw the line $PC$ intersect $\odot O$ at $L'$ and $D$, we will prove that $L'$ coincide with $L$. First, by power of point at $C$, we have \[ CK \cdot CO = CA \cdot CB = CD \cdot CL' \]Thus $K,D,O,L'$ concyclic. Moreover, $PA$ is tangent to $\odot O$ since $O \in \omega_2$, yielding $PL' \cdot PD = PA^2 = PS \cdot PO$, thus $D,O,S,L'$ concyclic as well, meaning that $K,D,O,S,L',$ lies on the same circle. Now remarking that $\angle CKP = \angle CSP = 90^\circ$ and $PK=PS$, we have $PC$ is the orthogonal bisector of $KS$. Thus $\angle L'SC = \angle L'KC =\angle L'DO =\angle L'SP$, implying that $L'S$ bisects the angle $\angle ASL$, thus $L'$ coincides with $L$ which means $L$ lies $PC$, the orthogonal bisector of $KS$.

Iran TST #3 2018 P1 wrote: Two circles $\omega_1(O)$ and $\omega_2$ intersect each other at $A,B$ ,and $O$ lies on $\omega_2$. Let $S$ be a point on $AB$ such that $OS\perp AB$. Line $OS$ intersects $\omega_2$  at $P$ (other than $O$). The bisector of $\hat{ASP}$ intersects  $\omega_1$ at $L$ ($A$ and $L$ are on the same side of the line $OP$). Let $K$ be a point on $\omega_2$ such that $PS=PK$ ($A$ and $K$ are on the same side of the line $OP$). Prove that $SL=KL$. Solution: Let $C \in (O)$, such, $\angle CSP=\angle ASC=45^{\circ}$. Let $CP \cap (O)=D$. Since, $\angle OAP=90^{\circ}$ $$-1=( D, ~C ; ~ B, ~A ) \overset{A}{=} ( D, ~C ; AB ~ \cap ~ DP , ~ P) \implies D \equiv L$$Let $L', K'$ be reflections of $L, K$ over $OP$. $$\angle LSO=\angle CSP=\angle L'SO \implies L' - S - C$$Let $C'$ be the reflection of $C$ over $OP$ $\implies$ $L - S - C'$ and $L' - C' - P$ $$\angle LOC=2\angle LL'S=\angle LSC \implies LOSC \text{ and similarly, } L'OSC' \text{ are cyclic}$$Also, $\angle OKP=\angle OK'P=90^{\circ}$ $\implies$ $OK, OK'$ are tangent to $\odot (P)$ with radius $PS=PK$. Let $LC', L'C$ $\cap$ $\odot (P)$ $=$ $D', D$, then, $D'$ is the reflection of $D$ over $OS$ and $\angle DSD'=90^{\circ}$ $\implies$ $D - P - D'$. Let $LP$ $\cap$ $\odot (AOP)$ $=$ $E$, then by Radical Axes Theorem, $EO$ $\cap$ $AB$ $=$ $G$ lies on tangent at $L$ to $(O)$. By some simple congruency, $E$ is the center of $\odot (LOSCG)$. Suppose, $PE$ $\cap$ $AB$ $=$ $M$ $\implies$ $M$ is the orthocenter WRT $\Delta OGP$ ($OP=OG$). Let $GP$ $\cap$ $\odot (AOP)$ $=$ $K'$ $\implies$ $O - M - K'$, but then, $PS$ $=$ $PK'$ $\implies$ $K' \equiv K$ $\implies$ $K$ lies on $\odot (LOSCG)$ $\implies$ $LP$ is the perpendicular bisector of $SK$

Let $\omega$ be the circumcircle of $\triangle KSO$ and $D=PK\cap AB.$ First, observe that $PK=PS\implies \triangle SDP\cong \triangle KOP$ by LL, so $ODKS$ is an isosceles trapezoid and $D\in\omega.$ Since $DS\cap OK$ is the radical center of $\omega,\omega_1,\omega_2,$ the perpendicular bisector of $OD$ is the radical axis of $\omega,\omega_1.$ Now note that since the the intersection of the angle bisector of $\angle ASP$ with $\omega$ lies on the perpendicular bisector of $SK,$ it must lie on $\omega_1$ as well so this point must be $L.$ $\blacksquare$

As $$OL^2=OA^2=OS \cdot OP$$we have the similarity $$\triangle OSL \sim \triangle OLP,$$so $\angle OLP = 135^\circ$. Let $K' = (OSL) \cap (OP)$. Then $\angle OK'L = 45^\circ$ and $\overline{K'L}$ thus bisects $\angle OK'P$. But as $$\angle OLP = 135^\circ= 90^\circ + \frac 12 \angle OK'P,$$$L$ is in fact the incenter of $\triangle OK'P$. Thus, $\angle K'PL = \angle PLS$, so $$\triangle K'LP \cong \triangle SLP.$$This means $PK' = PS$, so $K'=K$; but this congruence also implies $LK=LS$, so we are done.