Let $\measuredangle$ denote a directed angle modulo $180^\circ$ and $T$ be the midpoint of $AB.$ I claim that $TCPD$ is always a parallelogram, which suffices, as $CD$ then passes through the fixed midpoint of $TP.$ However, $ATPC$ and $BTPD$ cyclic implies that $\measuredangle TCA=\measuredangle TPA=\measuredangle BPT=\measuredangle BDT,$ so we're done by the Parallelogram Isogonality Lemma. $\blacksquare$

We claim that if $PO\cap AB=K$ then the desired fixed point is the midpoint of $PK$. Let $O$ be the center of $k$. Invert around $P$ with radius $PA$. It remains to show that $O$ is the midpoint of $PY'$. Let $C'D'\cap PY'=U$ and observe that the points $O, U$ are isogonal wrt $\sphericalangle PD'Y'$ and $\sphericalangle PC'Y'$ and consequently $(PC'Y'D')$ is harmonic. But then $D'U$ is the symmedian wrt $D$ in $\triangle D'PY'$ and hence $D'O$ is the respective median.