Let $O$ be the center of equilateral $\triangle APR$ and let $M$ be the midpoint of $PA.$ Since $\angle ABP=\angle AOP=120^{\circ}$ $\Longrightarrow$ $B \in \odot(POA)$ and obviously $RA,RP$ are tangents of $\odot(POA)$ $\Longrightarrow$ $BY$ is the B-symmedian of $\triangle PBA,$ isogonal of its B-median $BM$ $\Longrightarrow$ $\angle PBY=\angle ABM=\angle PCA.$ But as $\triangle PAC \cong \triangle PRQ$ are directly congruent, then $\angle PQR=\angle PCA$ $\Longrightarrow$ $\angle PQR=\angle PBY$ $\Longrightarrow$ $PQRB$ is cyclic $\Longrightarrow$ $\angle RBQ=\angle RPQ=\angle APC$ $\Longrightarrow$ $PXBY$ is cyclic $\Longrightarrow$ $\angle PYX=\angle PBX=\angle PRQ=\angle PAC$ $\Longrightarrow$ $XY \parallel AC.$

Very nice, as usual, Luis! A small remark for your proof: also $BQ$ is the $B-$symmedian of $\triangle BCP$; with $BC=AB$, done! Another idea: A $90^\circ$ rotation about $P$ sends $\triangle PAC$ to $\triangle PRQ$ and $B$ to $T$, midpoint of $QR$; since $\angle PBC=60^\circ$ and that is the rotation angle value, $T$ lies onto $AC$ and $QR\parallel PB$. As $\triangle PTQ\equiv\triangle BTR$ (s.a.s. criterion), $PBRQ$ is an isosceles trapezoid, i.e. $\triangle BRQ\cong\triangle PQR\cong\triangle PCA$, hence $\angle RBQ=\angle APC\ (\ 1\ )$ and $\angle CBQ=\angle APB\ (\ 2\ )$. From $(1)$ we infer $PYBX$ cyclic, i.e. $\angle YXB=\angle APB\ (\ 3\ )$; with $(2)$, done. Best regards, sunken rock