### Part (a)
1. **Identify the intersection point $X$**:
- Let $X$ be the intersection point of the perpendiculars from $M$ and $N$ to the lines $DM$ and $DN$, respectively.

2. Prove cyclic quadrilaterals:
- Since $O$ is a circle passing through $C$ and $D$, and intersects $BC$ at $M$ and $CA$ at $N$, the quadrilateral $MCND$ is cyclic.
- Therefore, $\angle MCD = \angle MND$ and $\angle NCD = \angle NMD$.

3. Angle relationships:
- In the cyclic quadrilateral $DMXN$, we have:
$$\angle MXD = \angle MCD = \frac{\angle C}{2}$$
$$\angle DXN = \angle DCN = \frac{\angle C}{2}$$
- Thus, $\angle MXD = \angle DXN$.

4. Congruence of triangles:
- The triangles $DMX$ and $DNX$ are congruent by the Angle-Side-Angle (ASA) criterion:
$$\angle MXD = \angle DXN, \quad DM = DN, \quad \text{and} \quad DX \text{ is common}.$$
- Therefore, $XM = XN$.

5. **Circle $S$**:
- The circle with center $X$ and radius $XM = XN$ is tangent to the lines $DM$ and $DN$ at points $M$ and $N$, respectively.

This completes the proof for part (a).

### Part (b)
1. Congruence of triangles:
- Consider the triangles $MXC$ and $QXC$:
$$XM = XQ, \quad \angle XMC = \angle XQC, \quad \text{and} \quad XC \text{ is common}.$$
- Therefore, $\triangle MXC \cong \triangle QXC$, which implies $MC = QC$.

2. Congruence of triangles:
- Similarly, consider the triangles $XCP$ and $XCN$:
$$XP = XN, \quad \angle XCP = \angle XCN, \quad \text{and} \quad XC \text{ is common}.$$
- Therefore, $\triangle XCP \cong \triangle XCN$, which implies $CP = CN$.

3. Equality of segments:
- Since $MP = NQ$, it suffices to prove that $MC + CN$ is constant.

4. Application of Ptolemy's theorem:
- Applying Ptolemy's theorem to the cyclic quadrilateral $MCND$:
$$MC \cdot ND + CN \cdot MD = CD \cdot MN$$
- Given $MD = DN$ from the congruence of triangles $DMX$ and $DNX$, we have:
$$MC + CN = CD \cdot \frac{MN}{MD}$$
- Since $\frac{MN}{MD} = 2 \cos \left( \frac{\angle C}{2} \right)$, the expression $MC + CN$ is constant.

This completes the proof for part (b).