Let the parallel line WRT $BC$ through $A$ intersect $(ABC)$ at $D$. By symmetry, $ADKL$ is cyclic. Applying a $\sqrt{bc}$ inversion through $A$ followed by a reflection through the internal angle bisector of $\angle BAC$ gives us that $P$ is the image of $K$. Hence, $\angle BAP = \angle CAK$. $\blacksquare$

Note if $M$ is midpoint of $BC$ then $K$ is reflection of $L$ above $M$. Take point $E$ on $(ABC)$ such that $AE \parallel BC$. If $O$ is circumcenter of $(ABC)$ then $A$ is reflection of $E$ and $L$ is reflection of $K$ above $OM$. $AL$ is reflection of $KE$ and $AE \parallel LK$. Hence $A,E,K,L$ cyclic. Take $\sqrt{bc}$ inversions. Its well know $T \leftrightarrow E$ (As $T'$ should lie on $(ABC)$ and $\measuredangle{TAI}=\measuredangle{IAT'}=\measuredangle{IAE}$) and $A_0 \leftrightarrow L$ Let $P'$ be image of $P$ then $P'$ should lie on $BC$ and $(AEL)$ hence $P'=K$ As $\sqrt{bc}$ inversions is reflection about angle bisector we get $AP$ is reflections of $AK$ hence $\measuredangle{BAP}=\measuredangle{KAC}$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15.741534557788974cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.081393817971838, xmax = 37.66014073981714, ymin = -18.28892983088801, ymax = 7.701611330231495; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); /* draw figures */ draw(circle((11.095712613319792,-6.177893877780872), 10.192679672815137), linewidth(0.4)); draw((xmin, -0.006085192697768812*xmin-8.692576064908721)--(xmax, -0.006085192697768812*xmax-8.692576064908721), linewidth(0.4)); /* line */ draw((xmin, 0.7609797307563224*xmin-1.813196043720326)--(xmax, 0.7609797307563224*xmax-1.813196043720326), linewidth(0.4)); /* line */ draw((xmin, -2.9954813531338718*xmin + 16.680825838893796)--(xmax, -2.9954813531338718*xmax + 16.680825838893796), linewidth(0.4)); /* line */ draw((xmin, -0.38657790910946643*xmin-12.105004285069663)--(xmax, -0.38657790910946643*xmax-12.105004285069663), linewidth(0.4) + yqqqyq); /* line */ draw((4.285776955438827,-13.761790979412739)--(4.923256615628692,1.9333024500850746), linewidth(0.4)); draw((4.923256615628692,1.9333024500850746)--(13.672198269465575,-8.775774025980523), linewidth(0.4)); draw((xmin, -0.006085192697768574*xmin + 1.963261415291745)--(xmax, -0.006085192697768574*xmax + 1.963261415291745), linewidth(0.4)); /* line */ draw(circle((11.121554363944778,-1.9312328584077858), 7.304349945571356), linewidth(0.4) + yqqqyq); draw((17.36642423119154,1.857583377373747)--(11.033689341740999,-16.370384840563304), linewidth(0.4)); draw((4.923256615628692,1.9333024500850746)--(1.22,-8.7), linewidth(0.4)); draw((4.923256615628692,1.9333024500850746)--(20.94,-8.82), linewidth(0.4)); /* dots and labels */ dot((4.923256615628692,1.9333024500850746),dotstyle); label("$A$", (4.007988805578389,2.4563547240191688), NE * labelscalefactor); dot((1.22,-8.7),dotstyle); label("$B$", (1.326425034986684,-8.41723902706144), NE * labelscalefactor); dot((20.94,-8.82),dotstyle); label("$C$", (21.069806642639893,-8.535109962032504), NE * labelscalefactor); dot((-8.968445578518395,-8.638001345363984),linewidth(4.pt) + dotstyle); label("$T$", (-8.83994310626758,-8.387771293318673), NE * labelscalefactor); dot((8.487801730534423,-8.744225974019479),linewidth(4.pt) + dotstyle); label("$L$", (8.604955269449881,-8.505642228289737), NE * labelscalefactor); dot((11.033689341740999,-16.370384840563304),linewidth(4.pt) + dotstyle); label("$A_0$", (11.581196377469245,-16.019914332695034), NE * labelscalefactor); dot((4.285776955438827,-13.761790979412739),linewidth(4.pt) + dotstyle); label("$P$", (4.391069344234347,-13.515156964559935), NE * labelscalefactor); dot((11.08,-8.76),linewidth(4.pt) + dotstyle); label("$M$", (11.198115838813289,-8.535109962032504), NE * labelscalefactor); dot((13.672198269465575,-8.775774025980523),linewidth(4.pt) + dotstyle); label("$K$", (13.791276408176696,-8.535109962032504), NE * labelscalefactor); dot((11.095712613319792,-6.177893877780875),linewidth(4.pt) + dotstyle); label("$O$", (10.196212891559245,-6.76704593746655), NE * labelscalefactor); dot((11.157735884898583,4.014597085001561),linewidth(4.pt) + dotstyle); label("$N$", (11.286519040041586,4.2538864823278875), NE * labelscalefactor); dot((7.089519596849112,-4.555697916144881),linewidth(4.pt) + dotstyle); label("$I$", (7.219971783539881,-4.321224036816982), NE * labelscalefactor); dot((17.36642423119154,1.857583377373747),linewidth(4.pt) + dotstyle); label("$E$", (17.474743126022442,2.1027419191059784), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

OK only inversion solutions here u can also do this easily by sine rule and length chasing

This is also easy to do by barycentric coordinates, by setting $ABC$ as the reference triangle and let $P'$ be the point where the isogonal of $AK$ meets $\omega$. Now one gets $P' = (-a^2bc: b^3(b+c): c^3(b+c))$, $A_0 = (-a^2: b(b+c): c(b+c))$ and $T = (0: -b^2: c^2)$ Now its easy to check that $P'$, $T$ and $A_0$ and collinear and hence, $P \equiv P'$

My solution: Claim: $A,A_0,K,T$ are concyclic. Proof: Let $M$ be mid-point of $BC$. It is well known that the angle subtended by the orthocentre and incentre at $A$ is $|\angle B - \angle C|/2$, and $AH || A_0M$, so $\angle LA_0M = |B-C|/2$, hence $\angle AA_0K = \angle LA_0K = |B-C|$. It is easy to observe that $\angle TAA_0 = \angle B+\angle A/2$ (or C instead of B, depending upon T is farther from which) and $\angle ALT = \angle B + \angle A/2$ (or $\angle C+\angle A/2$). Hence, $\angle ATL = \angle ALK = |\angle B-\angle C|$. Thus, the claim follows. Reflect the line $AK$ about $AA_0$, and let it intersect $\omega$ at $P'$. We want to show $P=P'$. For this, let $\angle BAP' = CAK = \alpha$ and $\angle P'AA_0 = \angle KAA_0 = \beta$. In the configuration where $AC>AB$, by simple angle chasing in $\Delta AP'A_0$, we have $\angle AA_0P' = \angle C+\alpha$. But, due to the claim, we also have $\angle AA_0T = \angle AKT = C+ \alpha$, so $T,P',A_0$ are collinear. Similar equality holds when $AC<AB$. $\blacksquare$

Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. It suffices to show that $P,L,A'$ are collinear. By Pascal's theorem on $PA_0A_0AAA'$, we have that $AA_0 \cap PA'$ lies on $BC$, so we are done.

Let the line passing through $A$ and parallel to $BC$ intersect $w$ again at $A'$. Notice that under $\sqrt{bc}$-inversion, $\phi(A')=T$ and $\phi(A_0)= L$. Since $AKLA'$ is concyclic and $K\in BC$, $\phi(K)=\phi(AKLA')\cap \phi(BC)=TA_0\cap w=P$. Thus $AP$ and $AK$ are isogonal lines and the result follows.