$\omega$ can be either internally tangent or externally tangent to $\omega_1$ and $\omega_2,$ both cases give the same result. Let $Q$ be the exsimilicenter of $\omega_1 \sim \omega_2.$ $A$ is the insimilicenter/exsimilicenter of $\omega \sim \omega_1$ and $B$ is the insimilicenter/exsimilicenter of $\omega \sim \omega_2.$ By Monge and d'Alembert theorem, $A,B,Q$ are collinear. Since $P,Q$ are harmonically separeted from $O_1,O_2$ and $PX \perp QX,$ it follows that $XP,AB$ bisect $\angle O_1XO_2$ or $\angle O_1XP=\angle O_2XP.$

Here's my trigonometric solution to this problem. Let $\angle APO_1=\angle PAO_1=x$ and $\angle BPO_2=\angle PBO_2=y$. Standard angle-chasing yields $\angle BAP=y$ and $\angle ABP=x$. Scale the diagram so that $XP=1$. From right-angled triangle $AXP$, $AX=\frac{1}{tany}$ and $AP=\frac{1}{siny}$. The Sine Rule in triangle $AO_1P$ gives $PO_1=AO_1=\frac{1}{2cosxsiny}$. Similarly, $PO_2=\frac{1}{2cosysinx}$. Thus $\frac{PO_1^2}{PO_2^2}=\frac{(tanx)^2}{(tany)^2}$. Applying the Cosine Rule in triangle $XAO_1$ gives $O_1X^2=\frac{1}{(tany)^2}+\frac{1}{4(cosx)^2(siny)^2} -\frac{cos(x+y)}{cosxsinytany}$. Similarly, $O_2X^2=\frac{1}{(tanx)^2}+\frac{1}{4(cosy)^2(sinx)^2} -\frac{cos(x+y)}{cosysinxtanx}$. Thus $\frac{O_1X^2}{O_2X^2}=\frac{\frac{1}{(tany)^2}+\frac{1}{4(cosx)^2(siny)^2} -\frac{cos(x+y)}{cosxsinytany}}{\frac{1}{(tanx)^2}+\frac{1}{4(cosy)^2(sinx)^2} -\frac{cos(x+y)}{cosysinxtanx}}$ $=\frac{(cosy)^2(sinx)^2(1+4sinxsinycosxcosy)}{(cosy)^2(sinx)^2(1+4sinxsinycosxcosy)}$ $=\frac{(tanx)^2}{(tany)^2}$ $=\frac{PO_1^2}{PO_2^2}$ The desired result then follows from the Angle Bisector Theorem.

You can also invert through P. Let the images be the same point (instead of $A'$, keep it as $A$ for simplicity). The new picture becomes two parallel lines ($\omega_1$ and $\omega_2$) with the circle $\omega$ tangent to both lines at points A, B. The centers $O_1$ and $O_2$ map to the reflections of $P$ across $\omega_1$ and $\omega_2$ respectively. Also, the perpendiculars to $AP$ and $BP$ and $A, B$ respectively meet at $X$. We now must prove that $\angle{XO_1P} = \angle{XO_2P}$, equivalent to $XO_1 = XO_2$. This is probably obvious, but I haven't found a good synthetic solution. So instead, let the lines $\omega_1$ and $\omega_2$ be the lines $y = 1$ and $y = -1$. Also, let $A$ be point (0,1), and $B$ is (0,-1). Let P be point $(p_1, p_2)$. We can now easily compute that $O_1=(p_1,2-p_2)$, and $O_2=(p_1, -2-p_2)$, and $X = (\frac{p^2_2-1}{p_1}, -p_2)$. Now its obvious that $XO_1 = XO_2$, so we're done.

Denotes $AB \cap {O_1}{O_2} = Q$, by Menelaus theorem $\frac{{{O_2}B}}{{BO}}\frac{{OA}}{{A{O_1}}}\frac{{{O_1}Q}}{{Q{O_2}}} = 1$, so $\frac{{{O_1}Q}}{{Q{O_2}}} = \frac{{A{O_1}}}{{{O_2}B}} = \frac{{{O_1}P}}{{P{O_2}}}$, hence $X(PQ;{O_1}{O_2}) = (PQ;{O_1}{O_2}) = - 1$. Because $PX \bot QX$, we have $\angle {O_1}XP = \angle {O_2}XP$.

Attachments:

Let $\Gamma$ be the $O$-excircle of $\Delta OO_1O_2$. A simple length chasing yields $\Gamma$ is tangent to $\Delta OO_1O_2$ at $A, P, B$ respectively. Letting $Q = AB \cap O_1O_2$, since $O_1$ is tangent to $\Gamma$ at $A, P$, we get $ -1 = (A, P; BO_1 \cap \Gamma, B) \overset{B}{=} (Q, P; O_1, O_2)$. But since $\angle PXQ = \angle PXB = 90$, a well known lemma yeilds $XP$ bisects $\angle O_1XO_2$, as desired.

Amazing Problem , but unncessarily made it look long. . ELMO Shortlist 2011 G2 wrote: Let $\omega,\omega_1,\omega_2$ be three mutually tangent circles such that $\omega_1,\omega_2$ are externally tangent at $P$, $\omega_1,\omega$ are internally tangent at $A$, and $\omega,\omega_2$ are internally tangent at $B$. Let $O,O_1,O_2$ be the centers of $\omega,\omega_1,\omega_2$, respectively. Given that $X$ is the foot of the perpendicular from $P$ to $AB$, prove that $\angle{O_1XP}=\angle{O_2XP}$. David Yang. Solution:- Claim 1:- $T(\text{defined later})$ is the $O-$ excenter of $\triangle OO_1O_2$. From Monge's Theorem or by Radical Axis Theorem we get that tangents to ($\omega,\omega_1$), ($\omega_1,\omega_2)$ and $(\omega_2,\omega)$ meet at a point $T$. Now notice that $T$ is the $O$-excenter of $\triangle OO_1O_2$ and $K,A,B$ are the tangency points made by the excircle with $O_1O_2,OO_1,OO_2$ respectively. Claim 2:- $OP,O_1B,O_2A$ concurs. Let us restate the problem. Restated Reduced Problem wrote: If $OO_1O_2$ is a triangle and the $O-$ excircle touches $O_1O_2,OO_1,OO_2$ at points $P,A,B$ respctively then $OP,O_1B,O_2A$ concurs. Proof:- It immediately follows from Brianchon's Theorem on $OBO_2PO_1A$. Claim 3:- $(O_1,O_2;P,K)$ is harmonic. Let $O_1O_2\cap AB=K$. Note that $P\in O_1O_2$. Now as $OP,O_1B,O_2A$ concurs, we can say that $(A,B;K,R)=-1$ where $R=OP\cap AB$. Note that $O-O_1-A$ and $O-O_2-B$. So, $$-1=(A,B;R,K)\overset{O}{=}(O_1,O_2;P,K)$$ Main Proof:- As $(O_1,O_2;P,K)$ is harmonic and $PX\perp AB$. So, $XP$ must be the internal bisector of $\angle O_1XO_2$. Hence, $\angle O_1XP=\angle O_2XP$. $\blacksquare$

$OP$,$AO_2$,$BO_1$ concurrent