I'm assuming here that $\omega, P$ are fixed otherwise the problem is not true. $\angle PX_2P=\angle PXO=\angle PX_1O$ this with $X_1O=X_2O$ gives that $X_1,X_2$ are symetric w.r.t. $OP$ therefore all such lines $X_1X_2$ are parallel because all go through $\infty_{\perp OP}$, thus done

You could also use spiral similarity to reach the same conclusion. (Very similar to MathLuis' solution) Extend $XO$ past $O$ so that it intersects $\omega$ at point $Q.$ Then by spiral similarity, $\triangle X_2PX_1$ is similar to $\triangle X_2OQ.$ Note that since $\triangle OX_2Q$ is isosceles, $PX_2 = PX_1$ as well. This would mean $PO$ makes a $90$ degree angle with segment $X_1X_2$ as the foot of the perpendicular has to be the midpoint of $X_1X_2.$

We claim that $PX_1=PX_2$ which follows from $$\angle PX_1X_2+\angle PX_2X_1=\angle XPX_2=\angle XOX_2=2\angle PX_1X_2$$Thus $PX_1=PX_2$ and $OX_1=OX_2$ so $X_1$ and $X_2$ are reflections about $OP$. So $X_1X_2\perp OP$, as desired.

Attachments:

Observe that $\angle PX_{2}X_{1}=\angle X_{2}PX-\angle X_{1}X_{2}P$ $=\angle X_{2}OX-X_{1}X_{2}P=2X_{2}X_{1}P-\angle X_{1}X_{2}P$ $\implies \angle PX_{1}X_{2}=\angle PX_{2}X_{1}$ and we are done.$\blacksquare$