The homothety from $k$ to $K$ maps $Q$ to the midpoint of arc $AB$ because it also maps $AB$ to a tangent which is parallel to $AB$. This means $PQ$ goes through the midpoint of arc $AB$ and we are finished.

Solution sketch: Let the perpendicular bisector of AB intersect the larger circle K at a new point $X$. It is well known (or can be proved by inversion), that $P$, $Q$ and $X$ are collinear. Hence $\angle QPA=\angle XBA$ and similarly $\angle QPB = \angle XAB$ because these angles are over the same arcs. since triangle $\triangle XAB$ is isoceless the result follows immidiately.

My proof of the Lemma by Inversion: Call the chord $AB$ and let the circle k be tangent to big circle K at $P$ and $AB$ is tangent to $k$ at $Q$. $M$ is the Midpoint of arc $AB$. Then let $\alpha$ be a circle centered at M going through $A$ and $B$. We now invert around $\alpha$. Therefore $AB$ is sent to big circle $K$ and vice versa. It also follows that $k = k*$, since it is tangent to $AB$ and $K$, but $AB$ and $K$ just swap. So $Q*$ ist the tangency point of $k*$ and $(MB*C*)$, but this was just the definition of $P$, so $P$ and $Q$ are inverses. Hence $M$, $P$ and $Q$ are collinear. And we are done!!