Let $C$ be the reflection of $O_1$ about $A$. We just need to show that $CP=CA$. Suppose $CP\neq CA$, then notice that $\angle PCO_2=\angle O_2CA$ and $O_2P=O_2A$. Therefore, the quadrilateral $CPO_2A$ is inscribed. Contradiction.

Another Solution: After inverting WRT $A$ the problem transforms to the following problem: Problem: Let $ABC$ be a isosceles triangle ($BA=BC$) and circumcenter $O$. Let $B'$ be the antipode of $B$ WRT $\odot(ABC)$. The perpendicular bisector of $BC$ cuts at $Q$. Let $\omega = \odot(Q,\ QB)$ . Assume that $BB'\cap \omega = P_1$ and perpendicular bisector of $AB$ cuts $\omega$ at $P_2$ ($P_2$ and $C$ are in the same side of $AB$). prove that $P_2$ is reflection of $P_1$ WRT $AB'$. Proof: By angle chasing it's easy to see $QO=QB'\Longrightarrow B'P_1=OB=OB'$. Since $OP_2$ is reflection of $OQ$ throw $OB'$ we conclude that $BCP_2P_1$ is isosceles trapezoid $\Longrightarrow \angle P_2QB'=\angle CQO=\angle BQO=\angle P_1QB'$. Hence $P_2$ is reflection of $P_1$ WRT $AB'$. [asy][asy] import graph; size(12.08653620925151cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -18.57802925285825, xmax = 23.508506956393262, ymin = -4.788379139833202, ymax = 14.570478552809805; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); draw((3.595079441575428,-1.3758045842231923)--(3.533114767949068,-0.37118584606632443)--(2.5284960297922003,-0.43315051969268437)--(2.59046070341856,-1.4377692578495522)--cycle, linewidth(0.4)); draw((0.259926525192123,2.0788763808430986)--(0.423628678091605,3.072002775099939)--(-0.5694977161652354,3.2357049279994206)--(-0.7331998690647175,2.2425785337425803)--cycle, linewidth(0.4)); draw((-0.2498813771566022,5.1747107179851)--(0.7432450171002383,5.011008565085619)--(0.9069471699997201,6.0041349593424584)--(-0.08617922425712025,6.167837112241941)--cycle, linewidth(0.4)); draw((6.499177247083074,-0.1882397327854548)--(5.4945585089262075,-0.25020440641181496)--(5.556523182552567,-1.2548231445686826)--(6.561141920709435,-1.1928584709423224)--cycle, linewidth(0.4)); /* draw figures */ draw(circle((2.395264716183147,1.726897558152264), 5.087190595272201), linewidth(0.8)); draw((-0.08617922425712025,6.167837112241941)--(-1.3802205138723147,-1.6826800447567818), linewidth(0.8) + red); draw((-1.3802205138723147,-1.6826800447567818)--(6.561141920709435,-1.1928584709423224), linewidth(0.8) + red); draw(circle((2.1440229556185844,5.800221368306379), 8.271281979830691), linewidth(0.8)); draw((6.561141920709435,-1.1928584709423224)--(6.170749946238608,5.136475161061309), linewidth(0.8) + blue); draw((-0.08617922425712025,6.167837112241941)--(6.170749946238608,5.136475161061309), linewidth(0.8) + blue); draw((-0.08617922425712025,6.167837112241941)--(2.395264716183147,1.726897558152264), linewidth(0.8)); draw((2.395264716183147,1.726897558152264)--(6.561141920709435,-1.1928584709423224), linewidth(0.8)); draw((2.395264716183147,1.726897558152264)--(2.1440229556185844,5.800221368306379), linewidth(0.8)); draw((2.395264716183147,1.726897558152264)--(2.59046070341856,-1.4377692578495522), linewidth(0.8)); draw((2.1440229556185844,5.800221368306379)--(-1.3802205138723147,-1.6826800447567818), linewidth(0.8)); draw((8.652193886678878,0.6955356069716321)--(-0.7331998690647175,2.2425785337425803), linewidth(0.8)); draw((9.946235176294072,8.546052763970359)--(6.170749946238608,5.136475161061309), linewidth(0.8)); draw((8.2632218626711,6.738453262020678)--(7.977011967803791,7.05537851101494), linewidth(0.8)); draw((8.139973154728889,6.627149414016727)--(7.85376325986158,6.94407466301099), linewidth(0.8)); draw((6.170749946238608,5.136475161061309)--(2.395264716183147,1.726897558152264), linewidth(0.8)); draw((4.487736632615642,3.328875659111631)--(4.201526737748333,3.6458009081058935), linewidth(0.8)); draw((4.364487924673425,3.2175718111076805)--(4.078278029806117,3.5344970601019434), linewidth(0.8)); draw((2.395264716183147,1.726897558152264)--(-1.3802205138723147,-1.6826800447567818), linewidth(0.8)); draw((0.712251402560178,-0.08070194379741424)--(0.42604150769286936,0.23622330519684842), linewidth(0.8)); draw((0.5890026946179616,-0.19200579180136465)--(0.302792799750653,0.124919457192898), linewidth(0.8)); draw((9.946235176294072,8.546052763970359)--(2.1440229556185844,5.800221368306379), linewidth(0.8)); /* dots and labels */ dot((-0.08617922425712025,6.167837112241941),linewidth(2.pt)); label("$A$", (-0.45282425631475803,6.88387329249567), NE * labelscalefactor); dot((-1.3802205138723147,-1.6826800447567818),linewidth(2.pt)); label("$B$", (-3.4420596353258577,-2.7006591925873873), NE * labelscalefactor); dot((6.561141920709435,-1.1928584709423224),linewidth(2.pt)); label("$C$", (7.233781003999498,-2.2736255670143795), NE * labelscalefactor); dot((2.395264716183147,1.726897558152264),linewidth(2.pt)); dot((6.170749946238608,5.136475161061309),linewidth(2.pt)); label("$B'$", (6.14247284975735,5.79256513825354), NE * labelscalefactor); dot((2.1440229556185844,5.800221368306379),linewidth(2.pt)); label("$Q$", (1.7772402327887609,6.1721505832073245), NE * labelscalefactor); dot((9.946235176294072,8.546052763970359),linewidth(2.pt)); label("$P_1$", (10.792394550441283,8.212422349833915), NE * labelscalefactor); dot((8.652193886678878,0.6955356069716321),linewidth(2.pt)); label("$P_2$", (9.036811867530004,-0.5654910647223497), NE * labelscalefactor); dot((2.59046070341856,-1.4377692578495522),linewidth(2.pt)); dot((-0.7331998690647175,2.2425785337425803),linewidth(2.pt) + uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Dadgarnia wrote: Let $C$ be the reflection of $O_1$ about $A$. We just need to show that $CP=CA$. Suppose $CP\neq CA$, then notice that $\angle PCO_2=\angle O_2CA$ and $O_2P=O_2A$. Therefore, the quadrilateral $CPO_2A$ is inscribed. Contradiction. You can end it in another way : Get $D$ be reflection of $O_2$ about $l$ and $P'$ be a point on $\omega_2$ such that $CP'=O_1A$ Because $C,D$ are belong to that circle and it's easy to see that $P'CDO_2$ is cyclic then we can see that $P=P'$ and it's obvious to see that $QO_1=PC$ and finished $\blacksquare$

My solution: assume that the perpendicular from $O_2$ intersect to line $x$ at $H$ and $x$ intersect $w_2$ in $T$ and let $B$ be the second intersection of $w_1,w_2$. and $R_i$ Be the Radius of circles. by angle chasing we get two result: 1)$O_1,O_2,T$ collinear. 2)$\angle O_2.O_3T=\angle ATP$ from $2$ we get $AP=2R_2.sin \angle ATP=2R_2.sin \angle O_2.O_3T=2R_2.\frac{O_2H}{R_3}=2R_2.\frac{\frac{R_1}{2}}{R_3} (*)$ now let $P'\in w_1$ such that$\angle P'AT=\angle TAP$. we need to prove that $AP'=AP$. because $\triangle AP'B \sim APT$ (by Easy angle chasing)We have: $AP'=PT.\frac{R_1}{R_2} (**)$ also notice that:$\triangle O_1O_2O_3 \sim \triangle O_2PT$ So $PT=\frac {R_2^2}{R_3}$ so if you put this in $(**)$ we need to prove: $AP=AP'=\frac {R_2^2}{R_3}.\frac{R_1}{R_2}=R_2.\frac{ R_1}{R_3}$. which according $(*)$ its always true.

Another solution by a friend of mine (Amirmohammad Derakhshandeh): P.S : sorry its in persian , it would take a long time to type it in LaTeX

another solution (farsi)

Invert WRT $\omega_1$ solution. Lemma; Let a circle $\Omega$ with center $O$ and a circle $\omega$ with center $O_1$ which passes through $O$. Then, the inversion of $O_1$ WRT. $\Omega$ is the reflection of $O$ across radical axis of $\Omega,\omega$. Proof; Obvious After invert about $\omega_1$, We get this problem; let $\omega_1$ be a circle with $O_1$ as its center and $\omega_2$ be a line and intersect $\omega_1$ at $A$. $O_2$ is the reflection of $O_1$ about $\omega_2$. $X$ is a circle tangent to $\omega_1$ at $A$ and pass through $O_1$. $\omega_3$ is a line passing through $O_2$ and center of $X$ and $P,Q$ is the intersection of $\omega_2,\omega_1$ and $\omega_3$ respectively. Prove that $P$ is the inversion of $Q$ WRT. $X$ Proof Let $O$ be the center of $X$ (midpoint of $AO_1$). So, $P$ is the centroid of $\triangle AO_1O_2$. So, $OO_2=3\cdot OP$. Obviously, the midpoint of $O_2O_1$ lies on $X$. So, $O_2$ always lies on circle $A$ radius $AO_1$. Hence, $O_2'$ the reflection of $O_2$ about $O$ lies on $\omega_1$. By power of point, $OP \cdot OQ=\frac{1}{3}OO_2' \cdot OQ = \frac{1}{3}\mathcal{P}(O,\omega_1)=\frac{1}{3}\cdot 3OO_1^2=OO_1^2$. finish

Draw a line $l$ through $O_1$ parallel to $AP$ and let it intersect $\omega_2$ at $K$. Let $KP$ and $O_1A$ intersect at $L$. Then: $\angle O_1LP=180-(\angle LO_1K + \angle LKO_1)=180-2\angle LKO_1=180-\angle PO_2O_1$ So $L$ lies on $\omega_3$ Reflect $P$ about $x$ to get $P'$. Obviously $P'$ lies on $\omega_3$ and $O_1L\parallel PP'$, so $O_1LPP'$ is an isosceles trapezoid. Therefore $O_1P'=PL$. Since $LPA$ is an isosceles triangle, $O_1P'=PL=LA=O_1A$ so $O_1P'$ is equal to the radius of $\omega_1$. So it lies on this circle. $\blacksquare$ [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.263636363636369, xmax = 14.590909090909085, ymin = -7.26, ymax = 10.267272727272722; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-0.92,0.36), 6.38), linewidth(1)); draw(circle((-7.3,0.36), 4.370210965669833), linewidth(1)); draw(circle((-4.11,3.8486505088552887), 4.7272383452642055), linewidth(1)); draw((-5.803233237894917,4.4659022326789914)--(-0.3505473008613137,6.7145356733158446), linewidth(1) + ffvvqq); draw((-5.803233237894917,4.4659022326789914)--(5.46,0.36), linewidth(1)); draw((3.6053718740672163,4.857266881273711)--(-7.3,0.36), linewidth(1) + ffvvqq); draw((-4.306466475789836,8.571804465357976)--(3.6053718740672163,4.857266881273711), linewidth(1)); draw((-7.3,0.36)--(-4.306466475789836,8.571804465357976), linewidth(1)); draw((-3.076890269378113,-0.764316720318428)--(-7.3,0.36), linewidth(1)); draw((-3.076890269378113,-0.764316720318428)--(-0.3505473008613137,6.7145356733158446), linewidth(1) + linetype("4 4") + blue); /* dots and labels */ dot((-0.92,0.36),dotstyle); label("$O_2$", (-0.7181818181818244,0.19454545454545524), NE * labelscalefactor); dot((-7.3,0.36),dotstyle); label("$O_1$", (-7.9363636363636425,0.02181818181818257), NE * labelscalefactor); dot((-5.803233237894917,4.4659022326789914),dotstyle); label("$A$", (-6.490909090909097,4.085454545454543), NE * labelscalefactor); dot((-4.11,3.8486505088552887),linewidth(4pt) + dotstyle); label("$O_3$", (-4.045454545454551,3.994545454545453), NE * labelscalefactor); dot((-0.3505473008613137,6.7145356733158446),linewidth(4pt) + dotstyle); label("$P$", (-0.281818181818188,6.867272727272724), NE * labelscalefactor); dot((3.6053718740672163,4.857266881273711),linewidth(4pt) + dotstyle); label("$K$", (3.6818181818181754,4.994545454545452), NE * labelscalefactor); dot((5.46,0.36),linewidth(4pt) + dotstyle); label("$l$", (-0.5727272727272789,3.2490909090909077), NE * labelscalefactor,ffvvqq); dot((-4.306466475789836,8.571804465357976),linewidth(4pt) + dotstyle); label("$L$", (-4.227272727272733,8.721818181818177), NE * labelscalefactor); dot((-3.076890269378113,-0.764316720318428),linewidth(4pt) + dotstyle); label("$P'$", (-3.009090909090915,-0.6236363636363624), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.5717845622174cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = 0.18137133989157075, xmax = 7.229227714703171, ymin = -3.715349796192219, ymax = 1.134969224112721; /* image dimensions */ pen qqttcc = rgb(0.,0.2,0.8); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pair O_1 = (3.671051240559019,-1.5144743797204903), O_2 = (1.6716319305596052,-1.4662828576292808), B = (5.670470550558433,-1.5626659018116995), A = (2.0724159715365196,-0.312656890815723), O_3 = (2.694518690037728,-0.5287831298187438), P = (3.6415260311705024,0.48530767390699436), G = (1.7179861395164364,0.4569081200830026), C = (3.2251709475820585,-0.7131383387428082); /* draw figures */ draw(circle(O_1, 2.), linewidth(1.2)); draw(O_2--B, linewidth(1.2)); draw(O_2--A, linewidth(1.2)); draw(A--B, linewidth(1.2)); draw(circle(O_3, 1.387516869216546), linewidth(1.2)); draw(O_1--P, linewidth(1.2)); draw(O_2--O_3, linewidth(1.2)); draw(O_1--O_3, linewidth(1.2)); draw(O_3--P, linewidth(1.2)); draw(G--(2.808815863993616,-1.9115843513926107), linewidth(1.2)); draw(G--O_3, linewidth(1.2)); draw((2.808815863993616,-1.9115843513926107)--O_2, linewidth(1.2)); draw(O_2--G, linewidth(1.2)); draw(P--G, linewidth(1.2)); draw(P--O_2, linewidth(1.2)); draw(O_3--(2.808815863993616,-1.9115843513926107), linewidth(1.2)); draw(B--P, linewidth(1.2)); draw(P--A, linewidth(1.2)); draw(O_1--(2.808815863993616,-1.9115843513926107), linewidth(1.2)); draw(P--C, linewidth(1.2) + linetype("4 4")); draw((3.4510452776616125,-0.12006341726112421)--(3.415651701090948,-0.10776724757468933), linewidth(1.2) + linetype("4 4")); draw(C--(2.808815863993616,-1.9115843513926107), linewidth(1.2) + linetype("4 4")); draw((3.0346901940731694,-1.3185094299109266)--(2.999296617502504,-1.306213260224492), linewidth(1.2) + linetype("4 4")); /* dots and labels */ dot(O_1,qqttcc); label("$O_1$", (3.7446416282731936,-1.6358387446130445), S * labelscalefactor,qqttcc); dot(O_2,qqttcc); label("$O_2$", (1.4852809722331108,-1.5009515412673682), SW * labelscalefactor,qqttcc); dot(B,linewidth(4.pt) + qqttcc); label("$B$", (5.6948857766461005,-1.5178124416855778), NE * labelscalefactor,qqttcc); dot(A,qqttcc); label("$A$", (2.04731098617343,-0.19704190892583157), NE * labelscalefactor,qqttcc); dot(O_3,linewidth(4.pt) + qqttcc); label("$O_3$", (2.7948109047140544,-0.5230193170112157), NE * labelscalefactor,qqttcc); dot(P,linewidth(4.pt) + qqttcc); label("$P$", (3.665957426321549,0.5279768090571779), NE * labelscalefactor,qqttcc); dot((2.808815863993616,-1.9115843513926107),linewidth(4.pt) + qqttcc); label("$P'$", (2.8903560070839087,-2.0573612550682827), NE * labelscalefactor,qqttcc); dot(G,linewidth(4.pt) + qqttcc); label("$G$", (1.6145478754393843,0.5504580096147906), NE * labelscalefactor,qqttcc); dot(C,linewidth(4.pt) + uuuuuu); label("$C$", (3.2500552160057126,-0.6691471206356983), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] easy solution without inversion and this kind of stuffs Let $P'$ be the reflection of $P$ about line $AB$,it is pretty obvious that $O_2,O_1,B$ lie on a single line.also we can easily observe that $PO_1P'O_2$ are on a circle centered at $O_3$ because $O_3P=O_3O_1=O_3P'$. $AP'$ intersects circle $(PGO_2)$ at $G$ then we know that $PP'O_2 \sim PAG$ so $G,O_3,O_1$ are collinear by a little bit of angle chasing we get that $\angle O_2AP'=\angle O_2P'A$ so we are done.

Here's another solution: Note that $x$ and $O_1O_2$ and $\omega_2$ are concurrent at $C$. now , define $Q = (O_1O_2P) \cap \omega_1$ . reflecting $O_1$ over $x$ (name it $O_1 '$), it's enough to prove that $O_1 ' P = O_1Q$ . now , looking the diagram from the triangle $CO_1O_1 '$ , the main problem is equivalent to this: "In an isosceles triangle $ABC$ with $AB=AC$ , let $M,N$ be midpoints of $BC , AC $ respectively . If $(AMB) \cap (BCN) = P$ , prove that $CP=CM$" Proof . By a homothety with center $C$ and radius 2 , we have that the reflection of $C$ over $P$ lies on $(AMB)$ . And by checking Power of $C$ with respect to $(AMB)$ we are done.

Synthetic. Let $P'$ be the reflection of $P$ over $x$. Let $C$ be the center of $\omega_3$. Let $X$ be the reflection of $O_1$ over the point $O_2$. Let $\alpha=\angle PXO_1$, we claim that $\angle O_1PA=\angle O_1AP=\alpha$. Firstly note that the reflection of $P$ over $x$ lies on $\omega_3$, because the center of $\omega_3$, $C$ lies on $x$. We have \begin{align*} \angle O_1AP'&=\angle O_1AP-\angle P'AP\\&=180^\circ-\angle PXO_1-2\angle XAP\\&=180^\circ-\angle PXO_1-\angle XO_2P\\&=\angle O_1O_2P-\angle PXO_1\\&=2\angle PXO_1-\angle PXO_1\\&=\alpha \end{align*}and \begin{align*} \angle O_1P'A&=\angle O_1P'P-\angle AP'P\\&=\angle O_1O_2P-\angle APP'\\&=2\angle PXO_1-90^\circ+\angle PAX\\ &=2\angle PXO_1-90^\circ+\frac{\angle{XO_2P}}{2}\\&=2\angle PXO_1-90^\circ+\frac{180^\circ-\angle{O_1O_2P}}{2}\\&=2\angle PXO_1-90^\circ+90^\circ-\angle PXO_1\\&=\alpha. \end{align*}We are done.

Let S be reflection of P across line and K be reflection of O1. note that we can prove KA = KP instead. O2O1 = O2P so ∠O1KO2 = ∠O2KP. we know O2 lies on perpendicular bisector of AP so KPO2A is cyclic or kite. it's not cyclic because KPO2O1 is cyclic and A lies between O1K so KPO2A is kite and KA = KP. we're Done.