IRL ABooklet

Solution 1. Draw the tangent to $\Omega_1$ at $A$ and let $H$ be its second point of intersection with $\Omega_2$. First suppose $P$ is not inside circle $\Omega_2$.

By the Alternate Segment Theorem, $\angle HAE = \angle PBA$. From the cyclic quadrilateral $ABFE$ we see that $\angle PBA = \angle AEF$, hence $\angle HAE = \angle AEF$. We also have $\angle FAH = \angle FEH$ (standing on the same arc of $\Omega_2$). Therefore,
$$\angle FAE = \angle FAH + \angle HAE = \angle FEH + \angle AEF = \angle AEH.$$
This implies that the segment $EF$ has the same length as $AH$. If $P$ is inside circle $\Omega_2$, we can proceed in a similar way.

Looking at triangle $PBE$ we find that $\angle FBE = \angle BPA - \angle BEA$. The Alternate Segment Theorem shows that the angle between $AB$ and the tangent $AH$ is equal to $\angle BPA$. From cyclic quadrilateral $ABEH$ we see that the same angle between $AB$ and the tangent $AH$ is equal to $\angle BEH$. Hence,
$$\angle FBE = \angle BPA - \angle BEA = \angle BEH - \angle BEA = \angle AEH.$$
Therefore, for each $P$ on $\Omega_1$, segments $EF$ and $AH$ have the same length.

Solution 2. The length of the chord EF of circle $\Omega_2$ is determined by angle $\angle FAE$. First suppose $P$ is not inside circle $\Omega_2$.

In this case $\angle FAE$ is an external angle of triangle $PFA$, hence is equal to $\angle APF + \angle AFP$. These two angles are the same for each point $P$ on $\Omega_1$, because $\angle APF$ stands on the arc $AB$ of circle $\Omega_1$, and $\angle AFP$ stands on the arc $AB$ of circle $\Omega_2$.

If $P$ is on the arc of $\Omega_1$ that is inside the circle $\Omega_2$, the angle $\angle FAE$ is an internal angle of triangle $PFA$ and so is equal to $180^\circ - (\angle APF + \angle AFP)$. This angle subtends the same chord as the angle $\angle APF + \angle AFP$, which does not change when $P$ varies, and which has the same value as in the other case. Hence, $\angle FAE$ is the same for each point $P$ on $\Omega_1$.