It suffices to prove that $\measuredangle PXC=\measuredangle XYC=\measuredangle BCP$. Claim: Given any point $P'$ on the tangent at $B$ to $(ABC)$, $P'\infty_{AC}\cap AB=Q'$, $Q'\infty_{BC}\cap AC=X'$. $R'$ is a point on $BP$ s.t. $\measuredangle BAP'=\measuredangle R'AC$, then $\measuredangle P'X'C=\measuredangle BCR'$. Pf: Let $P'$ be a moving point on $BP$, then $P'\mapsto Q'\mapsto X'$ is projective, and when $P'\in\ell_{\infty}$, $X'\in\ell_\infty$, since $P'X'$ is an envelope of a conic(by Steiner conic, when $P'=AC\cap BP$, $X'=A\neq P''$, so $P'X'$ doesnot pass through a fixed point), $P'\mapsto P'X'\cap\ell_\infty=\infty_{P'X'}$ is projective. Also $P' \mapsto R'\mapsto\infty_{CR'}$ is projective. So we need to show that $\infty_{P'X'}$ and $\infty_{CR'}$ satisfy the angle relation above for three cases. Take $P'=B,BP\cap AC$, these two cases are trivial. Take $P'$ on the $A-$median, then $P'X'//AB$, and $R'$ is on the symmedian, so $CR'$ is tangent to $(ABC)$, therefore $\measuredangle P'X'C=\measuredangle BAC=\measuredangle BCR'$. So the claim is true. Now take $P'=P$, then $R'=P$, so $\measuredangle PXC=\measuredangle BCP$ as desired.

Note that by Pappus theorem on the hexagon $QPBCAY$, it follows that $\overline{BP} \cap \overline{AY}$ lies on the line at infinity, so it follows that $\overline{BP} \parallel \overline{AY}$. Therefore, $\measuredangle QAY = \measuredangle QBG = \measuredangle ACB = \measuredangle AXY$ (where $G = \overline{QX} \cup \overline{BP}$) and hence $\overline{QA}$ is tangent to $(AXY)$. Now, $\measuredangle QAP = \measuredangle PAC = \measuredangle APQ$, so $\triangle AQP$ is an isosceles triangle, and hence $QA = QP$. Consequently, $QP^2 = QA^2 = QY \cdot QX$, so $\overline{QP}$ is tangent to $(PXY)$ as well. Finally, $\measuredangle YCX = \measuredangle CPQ = \measuredangle YXP$, so $\overline{PX}$ is tangent to $(XYC)$ as desired.

$AQPX \stackrel{\text{def}}{\sim} ABP'C \implies BP' \parallel AC$ so $P, P'$ are isogonal so $\measuredangle YXP = \measuredangle BCP' = \measuredangle PCA = \measuredangle YCX$.

Let $R \in CS$ be such that $PR \parallel BC$. Moreover, let $BP$ and $AC$ intersect at $S$. It suffices to prove that $\angle PXC=\angle XYP,$ or equivalently that $\angle PXC=\angle CPR$, that is $RP^2=RC \cdot RX$. Lemma: $SA=\dfrac{c^2b}{c^2-a^2}, SB=\dfrac{abc}{c^2-a^2}$ and $SC=\dfrac{a^2b}{c^2-a^2}$. Proof: Note that triangles $SBC$ and $SAB$ are similar, and so $\dfrac{SB}{SA}=\dfrac{BC}{BA}=\dfrac{SC}{SB},$ hence $SB^2=SA \cdot SC,$ or equivalently $SB=\sqrt{SA \cdot SC},$ which gives us $\dfrac{a}{c}=\dfrac{SB}{SA}=\sqrt{\dfrac{SC}{SA}},$ hence $\dfrac{SC}{SA}=\dfrac{a^2}{c^2}.$ Thus, $\dfrac{SC}{b}=\dfrac{SC}{SA-SC}=\dfrac{a^2}{c^2-a^2},$ which implies the desired expression for $SC$. Now, it is easy to obtain the other two expressions for $SA,SB$ $\blacksquare$ To the problem, note that $\dfrac{AX}{XC}=\dfrac{AQ}{QB}=\dfrac{SP}{PB}=\dfrac{AS}{AB}=\dfrac{cb}{c^2-a^2}$, and so $\dfrac{CX}{b}=\dfrac{XC}{AX+XC}=\dfrac{c^2-a^2}{c^2-a^2+cb},$ which implies that $CX=\dfrac{b(c^2-a^2)}{c^2-a^2+cb}, \,\,\, (1)$. Moreover, note that $\dfrac{CR}{CS}=\dfrac{BP}{BS}=\dfrac{BA}{BA+AS}=\dfrac{c(c^2-a^2)}{c(c^2-a^2)+c^2b},$ and so $CR=\dfrac{a^2b}{c^2-a^2+cb}, \,\,\, (2)$. Furthermore, $\dfrac{PR}{BC}=\dfrac{SP}{SB}=\dfrac{AS}{AS+AB}=\dfrac{c^2b}{c^2b+c(c^2-a^2)},$ and so $PR=\dfrac{abc}{c^2-a^2+cb}, \,\,\, (3)$. To sum up, note that from relations $(1), \, (2)$ and $(3)$ we have that $RC \cdot RX=CR(CR+CX)=\ldots=\dfrac{a^2b^2c^2}{(c^2-a^2+cb)^2}=RP^2,$ and so $RP$ is tangent to $(CXP),$ hence $\angle PXC=\angle CPR,$ as desired.

This is a longer approach, but I just wanted to showcase that this problem is a perfect example of working backward and reducing the problem to simpler and simpler ones. Removing point Y: Notice that $PX$ being tangent to the circumcircle of $\triangle XYC$ is equivalent to $\angle PXC = \angle XYC$, but $\angle PXC = \angle QPX$ and $\angle XYC = \angle PCR$ due to the parallel lines. Removing points X and S: Let $R = \overline{PQ}\cap\overline{BC}$ and $S = \overline{PX}\cap\overline{BC}$. Now $\angle QPY = \angle PCR \iff \triangle RPS \sim \triangle RCP \iff RP^2 = RS\cdot RC$. However, $RS$ is not particularly nice to work with, so we replace it by: \[RS = \frac{PR}{PQ}\cdot QX = \frac{PR}{PQ}\cdot \frac{AQ}{AB}\cdot BC = \frac{PR}{AB}\cdot BC\]where the last equality follows from $\angle QPA = \angle PAC = \angle QAP \Longrightarrow AQ=PQ$. Therefore we want to show that $\frac{RP^2}{RC} = \frac{PR\cdot BC}{AB}$ or $\frac{RP}{RC} = \frac{BC}{AB}$. By sine law applied on $\triangle ABC$ and $\triangle BPR$ we have \[\frac{BC}{AB} = \frac{\sin\alpha}{\sin\gamma} = \frac{\sin\alpha}{\sin(180^{\circ}-\gamma)} = \frac{PR}{BP}\]simplifying what we have to prove even further down to $RC = BP$. This can be shown by computing \[\frac{RC}{BR} = \frac{AQ}{BQ} = \frac{PQ}{BQ} = \frac{BP}{BR}\Longrightarrow BP = RC\]as $\angle PBC = \angle BAC = \angle BQP\Longrightarrow \triangle BPR\sim\triangle QPB$.

Let $AP$ meets $(ABC)$ again at point $D$, and $BD\cap PQ=E(.)$, $CD\cap PX = F(.)$. Applying Menelaus theorem to the triangles $\triangle ABD$ (w.r.t. $QEP$) and $\triangle ACD$ (w.r.t $XFP$), we find \[ \frac{BQ}{AQ}\cdot \frac{AP}{DP}\cdot \frac{DE}{BE}=1, \]\[ \frac{CX}{AX}\cdot \frac{AP}{DP}\cdot \frac{DF}{CF}=1. \]From $XQ\parallel BC$, we get $\frac{AQ}{BQ}=\frac{AX}{CX}$ that implies $\frac{BE}{ED}=\frac{CF}{FD}$. So $EF$ should be parallel to $BC$, and $DE=DF$, $DB=DC$ as $\triangle BDC$ is isosceles. On the other hand, having $\angle DPE = \angle DAC = \angle DAB = \angle DBP$ we can obtain that $DP$ is tangent to $(\triangle BEP)$, i.e. \[ DP^2 = DE\cdot DB = DF\cdot DC. \]Thus, $DP$ is tangent to $(\triangle CFP)$ and $\angle DPF = \angle PCF$. Hence \[ \angle PYX = \angle PCB = \angle PCE + \angle DCB = \angle DPF + \angle DPE = \angle EDF = \angle PXC \]and $PX$ is tangent to $(\triangle XYC)$.