Solution:

Denote the feet of the perpendiculars from $P$ to the lines $BC$ and $DC$ by $M$ and $N$ respectively and let $O = AC \cap BD$. Since the points $O$, $M$ and $N$ are midpoints of $CA$, $CX$ and $CY$ respectively, it suffices to prove that $M$, $N$ and $O$ are collinear. According to Menelaus's theorem for $\triangle BCD$ and points $M$, $N$ and $O$ we have to prove that

$$\frac{BM}{MC} \cdot \frac{CN}{ND} \cdot \frac{DO}{OB} = 1$$

Since $DO = OB$ the above simplifies to $\frac{BM}{CM} = \frac{DN}{CN}$. It follows from $BM = BC + CM$ and $DN = DC - CN = AB - CN$ that the last equality is equivalent to:
$$\frac{BC}{CM} + 2 = \frac{AB}{CN}$$
Denote by $S$ the foot of the perpendicular from $B$ to $AC$. Since $\angle BCS = \angle CPM = \varphi$ and $\angle BAC = \angle ACD = \angle CPN = \psi$ we conclude that $\triangle CBS \sim \triangle PCM$ and $\triangle ABS \sim \triangle PCN$. Therefore
$$\frac{CM}{BS} = \frac{CP}{BC} \text{ and } \frac{CN}{BS} = \frac{CP}{AB}$$
and thus,
$$CM = \frac{CP \cdot BS}{BC} \text{ and } CN = \frac{CP \cdot BS}{AB}$$
Now equality (1) becomes $AB^2 - BC^2 = 2CP \cdot BS$. It follows from
$$AB^2 - BC^2 = AS^2 - CS^2 = (AS - CS)(AS + CS) = 2OS \cdot AC$$
that
$$DC^2 - BC^2 = 2CP \cdot BS \Longleftrightarrow 2OS \cdot AC = 2CP \cdot BS \Longleftrightarrow OS \cdot AC = CP \cdot BS.$$
Since $\angle ACP = \angle BSO = 90^\circ$ and $\angle CAP = \angle SBO$ we conclude that $\triangle ACP \sim \triangle BSO$. This implies $OS \cdot AC = CP \cdot BS$, which completes the proof.