Answer: $65^{\circ}$.

Solution. Since $K S$ is a tangent to the circle, then

$$
\angle K S P=\angle S Q P=\angle S L P=\angle K L S=\angle S N M .
$$

But $\angle K S P$ and $\angle S N M$ are corresponding angles for the lines $P S$ and $M N$, so $P S \| M N$ and the inscribed quadrilateral $S Q M P$ is a trapezoid. Therefore, it is an isosceles trapezoid and, hence, $\angle P S Q=\angle S P M=\angle S L M$, the latter by the inscribed quadrilateral $S P L M$. From the parallelism of the lines $K N$ and $L M$ we have

$$
\angle S L M=\angle K S L=\angle K S P+\angle P S L=\angle S M P+\angle P M L=\angle L M S .
$$

Thus

$$
2 \angle P S Q=2 \angle S L M=\angle S L M+L M S=180^{\circ}-\angle L S M=130^{\circ}
$$

Comment. Correct solution - 20 points. Solution not completed, but with good progress - 10 points. Solution started, with some progress - 2 points. Incorrect or no solution - 0 points.
4. A circle passing through the vertices $L$ and $M$ of trapezoid $K L M N$ intersects the lateral sides $K L$ and $M N$ at points $P$ and $Q$ respectively and touches the base $K N$ at point $S$. It turns out that $\angle L S M=50^{\circ}$, and $\angle K L S=\angle S N M$. Find $\angle P S Q$.