### Part (a)

1. Angle Chasing and Cyclic Quadrilaterals:
- We start by noting that $ABCD$ is a cyclic quadrilateral, meaning that its opposite angles sum to $180^\circ$.
- Let $E$ be the intersection of the diagonals $AC$ and $BD$.
- The angle bisector of $\angle AEB$ intersects $AB, BC, CD, DA$ at $M, N, P, Q$ respectively.

2. Miquel Point:
- We need to show that the circles $(AQM), (BMN), (CNP), (DPQ)$ pass through a common point $K$.
- By definition, the Miquel point of a quadrilateral is the common point of the four circles passing through the vertices and the intersection points of the sides.

3. Angle Chasing:
- We use angle chasing to show that the angles subtended by the arcs in these circles are equal.
- Specifically, we need to show that the angles $\angle KAM$, $\angle KDP$, $\angle KBM$, and $\angle KCP$ are related in such a way that $K$ lies on all four circles.

4. **Proving the Existence of $K$:**
- By the properties of cyclic quadrilaterals and the angle bisectors, we can show that:
$$\angle KAM \equiv \angle KDP \quad \text{and} \quad \angle KBM \equiv \angle KCP \quad (\text{mod} \ \pi)$$
- This implies that $K$ is the Miquel point of the quadrilateral $ADPM$ and $BCPM$.

5. Conclusion:
- Therefore, the circles $(AQM), (BMN), (CNP), (DPQ)$ all pass through the common point $K$.

### Part (b)

1. **Distance from $O$ to $K$:**
- We need to prove that $OK \le \frac{2R^2}{\sqrt{4R^2 - m^2}}$, where $m = \min\{AC, BD\}$.

2. Using Power of a Point:
- Since $K$ is the Miquel point, we use the power of a point theorem:
$$\overline{OE} \cdot \overline{OK} = R^2$$
- This implies:
$$OK = \frac{R^2}{OE}$$

3. **Maximizing $OE$:**
- The maximum distance from $O$ to the diagonals $AC$ or $BD$ is given by:
$$OE \le \max(d(O, AC), d(O, BD))$$
- Since $m = \min\{AC, BD\}$, we have:
$$OE \le \frac{\sqrt{4R^2 - m^2}}{2}$$

4. Final Inequality:
- Substituting this into the expression for $OK$, we get:
$$OK \le \frac{R^2}{\frac{\sqrt{4R^2 - m^2}}{2}} = \frac{2R^2}{\sqrt{4R^2 - m^2}}$$

5. Conclusion:
- Therefore, we have shown that:
$$OK \le \frac{2R^2}{\sqrt{4R^2 - m^2}}$$

The final answer is $\frac{2R^2}{\sqrt{4R^2 - m^2}}$