Note that $AB$ is of a fixed length and thus by the Law of Sines, we are done.

why is AB of fixed length?

RV99 wrote: why is AB of fixed length? use the power theorem!

The angle $AB$ subtends is actually the sum of the measures of the arc $\overarc{XY}$ in both circles, from angle chasing, which is obviously constant.

XY IS A FIXED CHORD SO <PXY REMAINNS CONSTANT LIKEWISE <XAY &<XBY REMAINS CONSTANT ARE EQUAL BECAUSE XYAB IS A CYCLIC QUADRILATERAL LIKEWISE AFTER SOME CALCULATION WE GET<BXA &<AYB REMAINS CONSTANT and EQUAL SO AB IS A FIXED CHORD

I think it is clear enough $AA'BB'$ is an isosceles trapezium and hence $AB = A'B'$ that is length of $AB$ is constant.

Attachments:

Since ${XY}$ is fixed, ${\angle P}$ is also fixed. Therefore, by Sine Rule, the circumradius is also fixed. Now prove $\triangle PXY \sim \triangle PBA$. Therefore, the ratio of the circumradii of the two triangles is constant. Now, since the circumradius of $\triangle PXY$ is fixed the circumradius of $ \triangle PBA $ is also fixed. $\blacksquare$

This was such a beautiful question....though trivial

Claim: $\angle XPY$ is constant regardless of where $P$ is on $\omega_1$. Proof: On a circle, let $P, X, Y$ and $P'$ be four points. We are only required to prove $\angle XPY =\angle XP'Y$, which is obvious because of cyclic quadrilaterals. Now we know that $\angle XPY$ is constant. By law of sines, $\frac{AB}{\sin \angle XPY}=2R$ where $R$ is the circumradius of $\triangle ABP$. We are required to prove that $R$ is constant. We already know that $\sin \angle XPY$ is constant. Now, by Power of a Point, $$PX \cdot PA =PY \cdot PB \implies \frac{PA}{PB}=\frac{PX}{PY}$$so $\frac{PA}{PB}$ is constant. This implies $AB$ is constant, so $\frac{AB}{\sin \angle XPY}=2R$ is constant, implying the result.

Well this one was quite trivial. Consider two different points $P_1$ and $P_2$ on $\omega_1$. Now as $XY$ is a constant chord, we must have that $\angle XP_1Y=\angle XP_2Y$. Define two define chords $A_1B_1$ and $A_2B_2$ respectively for the points $P_1$ and $P_2$ respectively, Now notice that $\triangle P_1XY \sim \triangle P_1B_1A_1$ and $\triangle P_2XY \sim \triangle P_2B_2A_2$. The ratio of the circumradii of each of this pairs of triangles are individually equal, and also $\triangle P_1XY$ and $\triangle P_2XY$ have the same circumcricle, (and the same circumradius). This implies that $\triangle P_2A_2B_2$ and $\triangle P_1A_1B_1$ have the same circumradius too. Our proof is thus complete.