4. Let $S$ be the midpoint of arc $\overparen{A B C}$, $R$ and $T$ be the midpoints of $A C$ and $M N$ respectively. Then
$$\begin{array}{l} \triangle A M S \cong \triangle C N S \Rightarrow S M=S N \\ \Rightarrow \angle S R K=\angle S T K=90^{\circ}, \end{array}$$

i.e., $R$ and $T$ both lie on the circle $\Gamma_{1}$ with diameter $S K$.
Let $P Q$ intersect circle $\Gamma_{1}$ at point $D$. Then
$$D R=D T, \angle A R D=\angle D T N \text {. }$$

Let $P_{1}$ and $P_{2}$ be the points where the incircle of $\triangle A K M$ touches the lines $K A$ and $K M$ respectively, and $Q_{1}$ and $Q_{2}$ be the points where the excircle of $\triangle C K N$ touches the lines $K A$ and $K M$ respectively. Then
$$\begin{array}{l} R P_{1}+T P_{2}=R A+A P_{1}+M P_{2}+T M \\ =R A+A M+T M, \\ R Q_{1}+T Q_{2}=R C+C N+T N . \\ \text { Thus, } R P_{1}+T P_{2}=R Q_{1}+T Q_{2} . \\ \text { Also, } R P_{1}+R Q_{1}=P_{1} Q_{1}=P_{2} Q_{2}=T P_{2}+T Q_{2} \text {, so } \\ R P_{1}=T Q_{2}, R Q_{1}=T P_{2} . \\ \text { Hence, } R P_{1}=T Q_{2}, D R=D T, \\ \angle P_{1} R D=\angle Q_{2} T D . \end{array}$$
$$\text { Therefore, } \triangle D R P_{1} \cong \triangle D T Q_{2} \Rightarrow D P_{1}=D Q_{2} \text {. }$$

By symmetry, $D Q_{2}=D Q_{1}$, i.e., $\triangle D P_{1} Q_{1}$ is an isosceles triangle.
Thus, point $D$ lies on the midline of the right trapezoid $P Q Q_{1} P_{1}$.
Therefore, $D P=D Q$.
Since $S D \perp P Q$, it follows that $S P=S Q$.