Solution:

We consider the reflection $Z$ of $A$ under reflection across $BD$. We will prove that the two circles mentioned in the problem statement are tangent to each other at $Z$.

First, because of $\measuredangle BZR = \measuredangle RAB = \measuredangle BPR$ and the converse of the inscribed angle theorem, it is clear that $Z$ lies on the circumcircle $\omega_1$ of $BRP$. Analogously, one shows that $Z$ also lies on the circumcircle $\omega_2$ of $DQS$. The tangent-chord angle of the chord $ZB$ of $\omega_1$ is $\angle ZRB = \measuredangle BRA$. Since $AR$ is perpendicular to $BC$ and hence also to $AD$, this angle has the size $90^\circ - \angle ADB$. Analogously, one shows that the tangent-chord angle of the chord $ZD$ of $\omega_2$ equals $90^\circ - \angle DBA$. The sum of these two angles is now $180^\circ - \measuredangle ADB - \angle DBA = \angle BAD = \angle DZB$, which shows that the tangents to $\omega_1$ and $\omega_2$ at the point $Z$ coincide. Thus the two circles are indeed tangent to each other.