Solution:

Let $BY$ and $CX$ meet at $O$. $O$ is on the circumcircle of $AXY$, since $\triangle AXC \cong \triangle CYB$.

We claim that $KA$ and $KO$ are tangent to the circumcircle of $AXY$. Let $XY$ and $BC$ meet at $L$. Then, $LBZC$ is harmonic. A perspectivity at $X$ gives $AYOP$ is harmonic. Similarly, a perspectivity at $Y$ gives $AXOQ$ is harmonic. Thus, $K$ is the pole of chord $AO$.

Now we compute. Denote $r$ as the radius and $\theta$ as $\angle AXO$. Then,

$$\begin{gathered} r = \frac{XY}{\sqrt{3}} = \frac{\sqrt{5^2 + 3^2 - 3 \cdot 5}}{\sqrt{3}} = \sqrt{\frac{19}{3}} \\ \sin \theta = \sin 60^\circ \cdot \frac{AC}{XC} = \frac{\sqrt{3}}{2} \cdot \frac{8}{\sqrt{5^2 + 8^2 - 5 \cdot 8}} = \frac{4}{7} \sqrt{3} \\ KX \cdot KQ = KA^2 = (r \cdot \tan \theta)^2 = \left(\sqrt{\frac{19}{3}} \cdot 4 \sqrt{3}\right)^2 = 304. \end{gathered}$$