Just employ angle chasing. Firstly, $\triangle ABQ \equiv \triangle BQM$ and $\triangle CAP \equiv \triangle CPN$ due to two equal angles(the right angles and the bisected $\beta$ and $\gamma$ angles respectively) and the common side. Then $AQ=QM$ and $AP=PN$. So they are isosceles triangles. Then $$\angle PAN =\frac{\pi-\angle APN}{2}$$$$\angle MAQ =\frac{\pi-\angle AQM}{2}$$ Additionally $\angle APN=\pi-\angle BPN=\pi-(\frac{\pi}{2}-\beta)$ and $\angle AQM=\pi - \angle MQC=\pi-(\frac{\pi}{2}-\gamma)$ Finally $\angle PAN=45\deg-\beta/2$ and $\angle MAQ=45\deg -\gamma/2$ so $\angle MAN=\gamma/2+\beta/2=45\deg$

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It is technically British MO 1995.

Let angle ABC be 2x°, angle BCA be 2y° So, in ∆ ABC, 2x°+2y°=90° x°+y°=45° Since BP is the angle bisector of angle ABC So, angle ABP=angle PBC=x° Since CQ is the angle bisector of angle BCA So, angle BCQ=angle QCA=y° Observe that BMPA is cyclic , since angle PMB= angle BAP=90° SO, angle ABP=angle AMP=x Therefore, angle NMA=angle PMN- angle AMP angle NMA=90-x Observe that QNCA is cyclic , since angle QNC= angle QAC=90° SO, angle QNA=angle QCA=y Therefore, angle MNA=angle QNM- angle QNA angle MNA=90-y So, in ∆ MAN angle MAN=180°-(90°-x+90°-y) =x+y= 45° So, angle MAN is 45°