Problem 1542

1. Identify Key Points and Lines:
- Given quadrilateral $PQRS$ with an incircle touching sides $PQ, QR, RS,$ and $SP$ at points $A, B, C,$ and $D$ respectively.
- Line $RP$ intersects lines $BA$ and $BC$ at points $T$ and $M$ respectively.
- Point $N$ is on line $TB$ such that $NM$ bisects $\angle TMB$.
- Lines $CN$ and $TM$ intersect at $K$, and lines $BK$ and $CD$ intersect at $H$.

2. **Collinearity of Points $C, D, T$:**
- It is known that points $C, D, T$ are collinear. This can be shown using Pascal's theorem on the degenerate hexagon $PQCSDR$.

3. **Claim: $MH$ is the Angle Bisector of $\angle TMC$:**
- To prove $\angle NMH = 90^\circ$, we need to show that $MH$ is the angle bisector of $\angle TMC$.
- If $MH$ is the angle bisector of $\angle TMC$, then $\angle NMH$ will be half of $\angle BMC$, which is $180^\circ$.

4. **Using Ceva's Theorem in $\triangle BTC$:**
- Apply Ceva's theorem in $\triangle BTC$ with cevians $BM, CN,$ and $TH$.
- Ceva's theorem states that for cevians $AD, BE, CF$ of $\triangle ABC$ intersecting at a common point, $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$.

5. Verification:
- Verify that the cevians $BM, CN,$ and $TH$ intersect at a common point $K$.
- Check the ratios $\frac{BT}{TC}, \frac{CM}{MB}, \frac{NH}{HT}$ to ensure they satisfy Ceva's theorem.

6. Conclusion:
- Since $MH$ is the angle bisector of $\angle TMC$, $\angle NMH$ is half of $\angle BMC$, which is $90^\circ$.

$$\boxed{\angle NMH = 90^\circ}$$