1. Projection Setup:
We start by projecting all the points $A, M, O_A, O, O_B, N,$ and $B$ onto the line $CD$. Let $h_X$ denote the projection of point $X$ onto $CD$.

2. Collinearity:
- Let $O$ be the midpoint of $AB$. Since $O$ is the center of the semicircle, it lies on the perpendicular bisector of $AB$.
- $O_A$ is the circumcenter of $\triangle ACP$, so it lies on the perpendicular bisector of $CP$.
- $O_B$ is the circumcenter of $\triangle BDP$, so it lies on the perpendicular bisector of $DP$.
- $M$ is the midpoint of $AC$, and $N$ is the midpoint of $BD$.

3. Equal Projections:
- Since $O$ is the midpoint of $AB$, we have $AO = OB$.
- Since $P$ is the midpoint of $CD$, we have $CP = PD$.
- Therefore, the projections of $A$ and $B$ onto $CD$ are equidistant from $P$, i.e., $h_A P = h_B P$.
- Similarly, the projections of $C$ and $D$ onto $CD$ are equidistant from $P$, i.e., $h_C P = h_D P$.

4. Midpoints and Projections:
- Since $M$ is the midpoint of $AC$, the projection $h_M$ is the midpoint of $h_A$ and $h_C$.
- Since $N$ is the midpoint of $BD$, the projection $h_N$ is the midpoint of $h_B$ and $h_D$.
- Therefore, $h_M P = h_N P$.

5. Perpendicular Bisectors:
- Since $O_A$ lies on the perpendicular bisector of $CP$, the projection $h_{O_A}$ is equidistant from $h_C$ and $h_P$.
- Since $O_B$ lies on the perpendicular bisector of $DP$, the projection $h_{O_B}$ is equidistant from $h_D$ and $h_P$.
- Therefore, $h_{O_A} P = h_{O_B} P$.

6. Parallel Lines:
- We need to show that the lines $O_A O_B$ and $MN$ are parallel.
- Consider the ratios of the distances:
$$\frac{MO_A}{O_A O} = \frac{h_M h_{O_A}}{h_{O_A} P} = \frac{h_N h_{O_B}}{h_{O_B} P} = \frac{NO_B}{O_B O}$$
- Since the ratios are equal, the lines $O_A O_B$ and $MN$ are parallel by the properties of similar triangles and equal ratios.

$\blacksquare$