The key to getting started is to notice that angle $\mathrm{AQB}=90^{\circ}$.

Angle $\mathrm{BAQ}=90^{\circ}-\mathrm{B} / 2$, so angle $\mathrm{OAQ}=90^{\circ}-\mathrm{B} / 2-\mathrm{A} / 2=\mathrm{C} / 2$. So $\mathrm{OQ}=\mathrm{AO} \sin \mathrm{C} / 2$. Thus we have to show that $\mathrm{MP}=\mathrm{BC} \sin \mathrm{C} / 2$.

Let the incircle touch $\mathrm{AB}$ at $\mathrm{L}$ and let $\mathrm{Y}$ be the midpoint of ML (also the intersection of ML with AO). Angle NMC $=90^{\circ}$ - C/2. It is also A/2 + angle MPY, so angle MPY $=90-\mathrm{C} / 2-$ $\mathrm{A} / 2=\mathrm{B} / 2$. Hence $\mathrm{MP}=\mathrm{MY} / \sin \mathrm{B} / 2$. We have $\mathrm{MY}=\mathrm{MO} \sin \mathrm{MOA}=\mathrm{r} \cos \mathrm{A} / 2$ (where $\mathrm{r}$ is the inradius, as usual). So $\mathrm{MP}=(\mathrm{r} \cos \mathrm{A} / 2) / \sin \mathrm{B} / 2$. We have $\mathrm{BC}=\mathrm{BN}+\mathrm{NC}=\mathrm{r}(\cot \mathrm{B} / 2+$ $\cot \mathrm{C} / 2)$, so $\mathrm{MP} / \mathrm{BC}=(\cos \mathrm{A} / 2) /(\sin \mathrm{B} / 2(\cot \mathrm{B} / 2+\cot \mathrm{C} / 2))$. Hence $\mathrm{MP} /(\mathrm{BC} \sin \mathrm{C} / 2)=($ $\cos \mathrm{A} / 2) /(\cos \mathrm{B} / 2 \sin \mathrm{C} / 2+\sin \mathrm{B} / 2 \cos \mathrm{C} / 2)=\cos \mathrm{A} / 2 / \sin (\mathrm{B} / 2+\mathrm{C} / 2)=1$.