We use complex numbers with $(ABC)$ as the unit circle. Then $BP\parallel AC$ is equivalent to $\frac{b-p}{a-c}=\frac{\bar b-\bar p}{\bar a-\bar c}\iff p=b+ac\bar b-ac\bar p$ and $|a-b|=|a-p|\iff (a-b)(\bar a-\bar b)=(a-p)(\bar a-\bar p)\iff abc\bar p^2+(ab-b^2-bc-ac)\bar p+b-a+c=0$. Dividing out the factor of $\bar p-\bar b$ gives $\bar p=\frac{b+c-a}{abc}$ or $p=\frac{ac+ab-bc}{b}$. Similarly, $q=\frac{ab+ac-bc}{c}$. Now, $PQXO$ is cyclic if and only if $\frac{p(q-x)}{q(p-x)}\in\mathbb{R}$. Luckily, $p-x$ is a real multiple of $p-c$ and $q-x$ is a real multiple of $q-b$, so it remains to observe that \[\frac{p(q-b)}{q(p-c)}=\frac{\frac{ac+ab-bc}{b}\cdot\frac{ab+ac-2bc}{c}}{\frac{ab+ac-bc}{c}\cdot\frac{ac+ab-2bc}{b}}=1.\]

[asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.587324195155585, xmax = 22.449028776054927, ymin = -9.783507801641846, ymax = 8.209326405679441; /* image dimensions */ pen yqqqqq = rgb(0.5019607843137255,0,0); pen qqttzz = rgb(0,0.2,0.6); /* draw figures */ draw((5.90545454545454,2.0309090909090974)--(0.3418181818181788,-2.5872727272727243), linewidth(1)); draw((0.3418181818181788,-2.5872727272727243)--(6.4690909090909035,-3.0054545454545427), linewidth(1)); draw((6.4690909090909035,-3.0054545454545427)--(5.90545454545454,2.0309090909090974), linewidth(1)); draw((-0.5414273511274442,5.304953486467206)--(5.90545454545454,2.0309090909090974), linewidth(1)); draw((-0.5414273511274442,5.304953486467206)--(0.3418181818181788,-2.5872727272727243), linewidth(1)); draw((6.4690909090909035,-3.0054545454545427)--(10.751928593436,0.549580656583549), linewidth(1)); draw((10.751928593436,0.549580656583549)--(5.90545454545454,2.0309090909090974), linewidth(1)); draw((-0.5414273511274442,5.304953486467206)--(6.4690909090909035,-3.0054545454545427), linewidth(1)); draw(circle((6.1332513354398355,5.368627274606724), 6.674982391003855), linewidth(1) + linetype("0 3 4 3") + yqqqqq); draw((3.542850178018359,-0.7832189331459545)--(10.751928593436,0.549580656583549), linewidth(1)); draw((0.3418181818181788,-2.5872727272727243)--(4.945968740895384,-1.1999151787008084), linewidth(1)); draw((0.3418181818181788,-2.5872727272727243)--(0.9054545454545424,-7.623636363636364), linewidth(1) + linetype("0 3 4 3") + qqttzz); draw((0.9054545454545424,-7.623636363636364)--(6.4690909090909035,-3.0054545454545427), linewidth(1) + linetype("0 3 4 3") + qqttzz); draw((4.945968740895384,-1.1999151787008084)--(10.751928593436,0.549580656583549), linewidth(1)); /* dots and labels */ dot((5.90545454545454,2.0309090909090974),dotstyle); label("$A$", (5.727939394905569,2.062559745848478), NE * labelscalefactor); dot((0.3418181818181788,-2.5872727272727243),dotstyle); label("$B$", (-0.32462394446821324,-3.3305803503250497), NE * labelscalefactor); dot((6.4690909090909035,-3.0054545454545427),dotstyle); label("$C$", (6.905480900620312,-3.7780461224966526), NE * labelscalefactor); dot((-0.5414273511274442,5.304953486467206),linewidth(4pt) + dotstyle); label("$P$", (-1.1724538285828285,5.029964340249633), NE * labelscalefactor); dot((10.751928593436,0.549580656583549),linewidth(4pt) + dotstyle); label("$Q$", (10.932672850164735,-0.10411662466665113), NE * labelscalefactor); dot((4.945968740895384,-1.1999151787008084),linewidth(4pt) + dotstyle); label("$X$", (4.691702869876594,-2.2001405048388953), NE * labelscalefactor); dot((3.542850178018359,-0.7832189331459545),linewidth(4pt) + dotstyle); label("$O$", (3.3022038931331967,-1.5), NE * labelscalefactor); dot((0.9054545454545424,-7.623636363636364),linewidth(4pt) + dotstyle); label("$R$", (0.14639265781768418,-7.852339732269667), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $R$ be a point such that $BR \parallel AC$ and $CR \parallel AB$. Then note $AC \parallel PR$ and $AB \parallel QR$. If $O$ is the circumcenter of $ABC$, then $O$ is also the circumcenter of $RPQ$. Hence, we see $\angle XPR = \angle ARP$ and $\angle XQR = \angle ARQ$ which implies $\frac{1}{2} \angle PXQ = \angle BAC$, so $P, O, X, Q$ concyclic. $\quad \blacksquare$

Let me add another solution without adding new points (coz I am not smart enough to do so ) Notice that $A$ is the centre of spiral sim that takes $PB \longrightarrow CQ$ which means $\odot(QAXC)$ and $\odot(PAXB)$, also $PC=BQ$ because of the spiral sim $PC \longrightarrow BQ$. This implies $$\angle BXC=\angle PXQ=\angle PXA+\angle QXA=\angle PBA+\angle QCA=2 \cdot \angle BAC = \angle BOC \implies \odot(OXCB)$$Combining this with the fact that $OA=OB$ we get that $\triangle OBQ \cong \triangle OCP$ which also gives us that the spiral sim at $O$ takes $BQ \longrightarrow CP$ which means $\odot(POXQ)$ as desired $\blacksquare$ Not that bad