I solve this problem, Yeah

Lets call line $XZ$ as $l'$ and lets observe its movement while $l$ is changing. We have a similar situation in both of $\omega_1$ and $\omega_2$, it comes from angle chasing. Thus the $l'$ lines passing through a common point. Lets call it $C'$. If we take $AC$ as $l_0$, we get that $C'$ lies on the line passing through $A$ and tangent to $\omega_2$. Lets $\angle{CAC'} = \alpha$. Now we have below. There are two points $C$ and $C'$ on the plane and there are two lines passing through this points respectively and rotating. And always these lines intersect with angle $\alpha$. So their intersecting point move on a circle which also include $A$. Hence it is sufficient to prove that diameter of this circle is $AC$ i.e. this circle passing through $B$. And finally it comes from taking $BC$ as $l_0$.

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); pair A = dir(140), B = dir(220), P = dir(85), Q = dir(5), C = extension(P,Q,B,B+dir(0)), Y = foot(A,P,Q), O = extension((A+B)/2,origin,A,A+dir(B--foot(B,A,C))), X = intersectionpoint(circle(O,abs(O-A)),A+dir(P--A)*0.00069--A+dir(P--A)*69), Z = intersectionpoint(circle(O,abs(O-A)),A+dir(Q--A)*0.00069--A+dir(Q--A)*69); draw(circle(O,abs(O-A))^^unitcircle); draw(P--X--Y--C--B--A--C^^A--Y^^Z--Q^^P--B--Y^^B--X); draw(circumcircle(B,X,Y), dashed); draw(circumcircle(A,B,C), dotted); dot("$A$", A, dir(95)); dot("$B$", B, dir(260)); dot("$C$", C, dir(330)); dot("$P$", P, dir(45)); dot("$Q$", Q, dir(20)); dot("$Y$", Y, dir(75)); dot("$X$", X, dir(155)); dot("$Z$", Z, dir(95)); [/asy][/asy] Redefine $Y = \overline{XZ} \cap \overline{PQ}$; it suffices to show $\overline{AY} \perp \overline{PQ}$. Note that $BXYP$ is cyclic from $$\measuredangle YXB = \measuredangle ZXB = \measuredangle ZAB = \measuredangle QAB = \measuredangle APB = \measuredangle YPB.$$Then $$\measuredangle CYB = \measuredangle PYB = \measuredangle PXB = \measuredangle AXB = \measuredangle CAB,$$so $ABCY$ is also cyclic. It follows that $\angle AYC = 180^\circ - \angle ABC = 90^\circ$ as desired. $\blacksquare$

Claim. $ABCY$ is cyclic. $\measuredangle AYC=\measuredangle ABC=90^\circ$ Claim. $QBXY$ is cyclic. $$\measuredangle YQB=\measuredangle PQB=\measuredangle PAB=\measuredangle ZAB=\measuredangle ZXB=\measuredangle YXB$$ $$\measuredangle AXZ=\measuredangle CAZ=\measuredangle CAP=\measuredangle ACP+\measuredangle CPA =\measuredangle ACY+\measuredangle QPA=\measuredangle QBA+\measuredangle ABY=\measuredangle QBY=\measuredangle QXY$$We are done.

My glance solution during the exam let $XZ \cap CP = Y$ $\angle YXB = \angle ZXB = \angle QAB = \angle QPB \implies (YPBX)$ is cyclic. so $\angle CYB = \angle PYB = \angle PXB = \angle AXB = \angle CAB \implies (CYAB)$ is cyclic $\implies \angle CYA = \angle ABC = 90$

Slowly realising the importance of using directed angles which I avoided till now [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -13.730116838164975, xmax = 12.772716605435255, ymin = -8.97763067882407, ymax = 10.345252639846398; /* image dimensions */ pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-0.7429610189117499,-3.3522232522245514)--(-0.7651068710359651,-2.818262151007362)--(-1.2990679722531546,-2.8404080031315773)--(-1.2769221201289394,-3.3743691043487667)--cycle, linewidth(1)); draw((4.050974931125255,-0.8808125733708889)--(3.745951558585542,-1.319635588473653)--(4.184774573688306,-1.624658961013366)--(4.489797946228019,-1.185835945910602)--cycle, linewidth(1)); draw(arc((-11.431818091643693,1.3924841042795622),0.7557842236768126,-25.146031123123784,9.268725691088449)--(-11.431818091643693,1.3924841042795622)--cycle, linewidth(1) + blue); draw(arc((-1.5415677856032433,3.0065319409761027),0.7557842236768126,-87.62503848423863,-53.210281670026404)--(-1.5415677856032433,3.0065319409761027)--cycle, linewidth(1) + blue); draw(arc((4.489797946228019,-1.185835945910602),0.7557842236768126,-159.21770003771,-124.80294322349776)--(4.489797946228019,-1.185835945910602)--cycle, linewidth(1) + blue); draw(arc((-0.4163502877927109,-0.39134245160123865),0.7557842236768126,-106.09229472332194,-71.67753790910972)--(-0.4163502877927109,-0.39134245160123865)--cycle, linewidth(1) + blue); /* draw figures */ draw(circle((-6.066206992008649,-0.37706447271067817), 5.649874745412944), linewidth(1) + blue); draw(circle((4.121049652496891,0.04544847567895349), 6.390090104116835), linewidth(1) + blue); draw((-1.2769221201289394,-3.3743691043487667)--(-1.5415677856032433,3.0065319409761027), linewidth(1)); draw((4.489797946228019,-1.185835945910602)--(-11.431818091643693,1.3924841042795622), linewidth(1) + linetype("2 2") + sexdts); draw((3.0945863663053452,-3.1930623007639345)--(-1.5415677856032433,3.0065319409761027), linewidth(1)); draw((-1.2769221201289394,-3.3743691043487667)--(-11.431818091643693,1.3924841042795622), linewidth(1)); draw((-1.2769221201289394,-3.3743691043487667)--(4.489797946228019,-1.185835945910602), linewidth(1)); draw((-1.2769221201289394,-3.3743691043487667)--(3.0945863663053452,-3.1930623007639345), linewidth(1)); draw((-1.2769221201289394,-3.3743691043487667)--(-0.4163502877927109,-0.39134245160123865), linewidth(1)); draw(circle((1.7387378816364771,-2.628709024732895), 3.106479390013805), linewidth(1) + linetype("2 2") + dtsfsf); draw((-1.5415677856032433,3.0065319409761027)--(4.489797946228019,-1.185835945910602), linewidth(1)); draw((1.3449849897196908,-5.7101328613200835)--(8.548467248051384,4.65318381674471), linewidth(1)); draw((1.3449849897196908,-5.7101328613200835)--(-1.5415677856032433,3.0065319409761027), linewidth(1)); draw((-11.431818091643693,1.3924841042795622)--(8.548467248051384,4.65318381674471), linewidth(1)); draw(circle((0.7765092903510512,-0.09326517989391579), 3.8706877322536455), linewidth(1) + linetype("2 2") + ffvvqq); /* dots and labels */ dot((-1.5415677856032433,3.0065319409761027),dotstyle); label("$A$", (-1.813918911527229,3.492809011843312), NE * labelscalefactor); dot((-1.2769221201289394,-3.3743691043487667),dotstyle); label("$B$", (-1.5367980295123975,-3.9894548025571175), NE * labelscalefactor); dot((3.0945863663053452,-3.1930623007639345),dotstyle); label("$C$", (3.1994497721956283,-2.931356889409582), NE * labelscalefactor); dot((8.548467248051384,4.65318381674471),dotstyle); label("$P$", (8.741867412492255,4.752449384637997), NE * labelscalefactor); dot((1.3449849897196908,-5.7101328613200835),dotstyle); label("$Q$", (1.1588323682682342,-6.2820002810434445), NE * labelscalefactor); dot((-0.4163502877927109,-0.39134245160123865),dotstyle); label("$Z$", (-0.32754327162949737,-0.13495526180538112), NE * labelscalefactor); dot((-11.431818091643693,1.3924841042795622),dotstyle); label("$X$", (-12.067391546075987,1.3514203780923473), NE * labelscalefactor); dot((4.489797946228019,-1.185835945910602),dotstyle); label("$Y$", (4.836982256828723,-1.344210019688279), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Observe that $ABCE$ is cyclic because $$\measuredangle ABC = \measuredangle AYC = 90^{\circ}$$Now, the most important claim Claim: $BQYZ$ is cyclic $$\measuredangle BZQ = \measuredangle BXA = \measuredangle BAC = \measuredangle BYC = \measuredangle BYQ \square$$Now for the finish notice that $$\measuredangle XZY = \measuredangle XZB + \measuredangle BZY = \measuredangle XAB + \measuredangle BQY = \measuredangle XAB + \measuredangle BQP = \measuredangle XAB + \measuredangle BAP = 180^{\circ} \blacksquare$$

\begin{align*} \angle{BAC} &=\angle{BAX}+\angle{XAC} \\ &=\angle{BAX}+\angle{ABX} \\ &=\pi-\angle{AXB} \end{align*}But now notice, $$\angle{BYC}=\pi-\angle{BAC}$$$$\implies \angle{AXB}=\angle{BYC}$$This in fact proves that $BXPY$ and $BZQY$ are cyclic. Now it suffices to prove that $\angle{YBQ}=\angle{PAC}$. To finish, observe $$\angle{BZY}=\angle{BQY}=\angle{BAP}=\angle{BAX}=180-\angle{AXB}-\angle{ABX} \implies \angle{BZY}+\angle{AZB}+\angle{AZX}=\pi \blacksquare$$

There was literally no way i couldve solved this on paper, ggb orsmax. BTW A VERY CLEAN AND UNDERSTANDABLE DIAGRAM IS ATTACHED

Since $ABCY$ is cyclic, $$\measuredangle BYZ=\measuredangle BZA=\measuredangle BXA=\measuredangle BAC=\measuredangle BYC$$and $BQYZ$ is cyclic. Hence, $$\measuredangle XZB=\measuredangle XAB=\measuredangle PAB=\measuredangle PQB=\measuredangle YZB.$$$\square$