Case 1: $X$ and $P$ lie on the same side of the line $AO$.

Step 1: $\angle QAO = 90^\circ$ (angle in a semicircle) and similarly $\angle PAO = 90^\circ$.
Therefore $PAQ$ is a straight line and is a tangent to the circumcircle of $\triangle ABC$ at the point $A$.

Step 2: Note also that $CX \perp OQ$ since $OA = OC$ and $OQ$ is a diameter.

Step 3: Now
$$\begin{aligned} \angle AXO &= \angle APO \text{ (subtending same arc)} \\ &= 90^\circ - \angle KPQ \text{ since } KP \text{ is a tangent at } P. \end{aligned}$$
Also $\angle AXO = \angle CXO = 90^\circ - \angle XOQ$ since $OQ \perp CX$, and therefore $\angle KPQ = \angle XOQ$.

Step 4: Now $\angle KPO + \angle KQO = 90^\circ + 90^\circ = 180^\circ$, so $KPOQ$ is a cyclic quadrilateral.

Step 5: Therefore $\angle KOQ = \angle KPQ$.

Step 6: Therefore $\angle KOQ = \angle XOQ$.

Step 7: Therefore $KO$ passes through $X$.

Case 2: $X$ and $P$ lie on opposite sides of the line $AO$.

We have
$$\angle AXO + \angle APO = 180^\circ$$

and so
$$\begin{aligned} \angle AXO &= 180^\circ - \angle APO \\ &= 180^\circ - (90^\circ - \angle KPQ) = 90^\circ + \angle KPQ. \end{aligned}$$
Also
$$\begin{aligned} \angle AXO &= 180^\circ - \angle CXO \\ &= 180^\circ - (90^\circ - \angle XOQ) \\ &= 90^\circ + \angle XOQ. \end{aligned}$$
Therefore $\angle KPQ = \angle XOQ$.

Step 4: Now $\angle KPO + \angle KQO = 90^\circ + 90^\circ = 180^\circ$, so $KPOQ$ is a cyclic quadrilateral.

Step 5: Therefore $\angle KOQ = \angle KPQ$.

Step 6: Therefore $\angle KOQ = \angle XOQ$.

Step 7: Therefore $KO$ passes through $X$.