CroatianCompetitions

Since quadrilaterals $BFSD$ and $CDSE$ are cyclic, we have $\angle FSD = 180^\circ - \beta$ and $\angle ESD = 180^\circ - \gamma$ so $\angle FSP = \beta$ and $\angle ESP = \gamma$.

Let $x = \angle FAP$ and $y = \angle BAM$. Then $\angle EAP = \alpha - x$ and $\angle CAM = \alpha - y$.
Applying the law of sines on triangles $EAP$ and $FAP$ we get
$$\frac{|EP|}{\sin(\alpha - x)} = \frac{|AE|}{\sin(\angle APE)}, \quad \frac{|FP|}{\sin x} = \frac{|AF|}{\sin(\angle APF)}.$$
Dividing those two equalities because of $|AE| = |AF|$ and $\angle APE + \angle APF = 180^\circ$ we get
$$\frac{\sin x}{\sin(\alpha - x)} = \frac{|FP|}{|EP|}.$$
Applying the law of sines on triangles $EPS$ and $FPS$ we get
$$\frac{|EP|}{\sin(\angle PSE)} = \frac{|SP|}{\sin(\angle PES)}, \quad \frac{|FP|}{\sin(\angle PSF)} = \frac{|SP|}{\sin(\angle PFS)}$$
i.e.
$$\frac{|EP|}{\sin \gamma} = \frac{|SP|}{\sin \frac{\alpha}{2}}, \quad \frac{|FP|}{\sin \beta} = \frac{|SP|}{\sin \frac{\alpha}{2}} \quad \text{so} \quad \frac{|EP|}{|FP|} = \frac{\sin \gamma}{\sin \beta}.$$

Observing the triangles $ABM$ and $ACM$ we see that:
$$\frac{|BM|}{\sin y} = \frac{|AM|}{\sin \beta}, \quad \frac{|CM|}{\sin(\alpha - y)} = \frac{|AM|}{\sin \gamma}$$
so
$$\frac{\sin y}{\sin(\alpha - y)} = \frac{\sin \beta}{\sin \gamma}.$$
From all of the above we obtain the equation
$$\frac{\sin x}{\sin(\alpha - x)} = \frac{\sin y}{\sin(\alpha - y)}$$
from which we can deduce that $x = y$.