For brevity, we let $\ell_{i}$ denote line $U_{i} V_{i}$ for $i=1,2$. We first give an explicit description of the fixed point $K$. Let $E$ and $F$ be points on $\Gamma$ such that $\overline{A E} \| \ell_{1}$ and $\overline{B F} \| \ell_{2}$. The problem conditions imply that $E$ lies between $U_{1}$ and $A$ while $F$ lies between $U_{2}$ and $B$. Then we let
$$K=\overline{A F} \cap \overline{B E}$$
This point exists because $A E F B$ are the vertices of a convex quadrilateral. Remark (How to identify the fixed point). If we drop the condition that $X$ lies on the arc, then the choice above is motivated by choosing $X \in\{E, F\}$. Essentially, when one chooses $X \rightarrow E$, the point $C$ approaches an infinity point. So in this degenerate case, the only points whose power is finite to $(X C D)$ are bounded are those on line $B E$. The same logic shows that $K$ must lie on line $A F$. Therefore, if the problem is going to work, the fixed point must be exactly $\overline{A F} \cap \overline{B E}$.

【First approach by Vincent Huang. We need the following claim: Claim - Suppose distinct lines $A C$ and $B D$ meet at $X$. Then for any point $K$
$$\operatorname{pow}(K, X A B)+\operatorname{pow}(K, X C D)=\operatorname{pow}(K, X A D)+\operatorname{pow}(K, X B C)$$
Construct the points $P=\ell_{1} \cap \overline{B E}$ and $Q=\ell_{2} \cap \overline{A F}$, which do not depend on $X$. Claim - Quadrilaterals $B P C X$ and $A Q D X$ are cyclic. !
Now, for the particular $K$ we choose, we have
$$\begin{aligned} \operatorname{pow}(K, X C D) & =\operatorname{pow}(K, X A D)+\operatorname{pow}(K, X B C)-\operatorname{pow}(K, X A B) \\ & =K A \cdot K Q+K B \cdot K P-\operatorname{pow}(K, \Gamma) . \end{aligned}$$