1. Identify the given conditions and setup:
- $ABCDEF$ is a cyclic hexagon.
- $AB \perp BD$.
- $BC = EF$.
- $P$ is the intersection of lines $BC$ and $AD$.
- $Q$ is the intersection of lines $EF$ and $AD$.
- $P$ and $Q$ are on the same side of $D$.
- $A$ is on the opposite side.
- $S$ is the midpoint of $AD$.
- $K$ and $L$ are the incenters of $\triangle BPS$ and $\triangle EQS$ respectively.

2. **Prove that $S$ is the circumcenter of the hexagon:**
- Since $S$ is the midpoint of $AD$ and $AB \perp BD$, $S$ lies on the perpendicular bisector of $AD$.
- Given that $BC = EF$, triangles $\triangle BCS$ and $\triangle EFS$ are congruent by the Side-Angle-Side (SAS) criterion.
- Therefore, $\angle CBS = \angle FES$.

3. **Calculate the angles involving $K$ and $L$:**
- Since $K$ is the incenter of $\triangle BPS$, $\angle KBS = \frac{1}{2} \angle CBS$.
- Similarly, since $L$ is the incenter of $\triangle EQS$, $\angle LES = \frac{1}{2} \angle QES$.

4. Use the cyclic nature of the hexagon:
- Since $ABCDEF$ is cyclic, $\angle CBS + \angle QES = 180^\circ$.

5. **Sum the angles to find $\angle KDL$:**
- $\angle KDS = \angle KBS = \frac{1}{2} \angle CBS$.
- $\angle LDS = \angle LES = \frac{1}{2} \angle QES$.
- Therefore, $\angle KDS + \angle LDS = \frac{1}{2} \angle CBS + \frac{1}{2} \angle QES = \frac{1}{2} ( \angle CBS + \angle QES ) = \frac{1}{2} \times 180^\circ = 90^\circ$.

Thus, $\angle KDL = 90^\circ$.

$\blacksquare$