1. Identify the given elements and their properties:
- Let $\triangle ABC$ be a triangle with circumcircle $\Omega$.
- The angle bisectors of $\angle BAC$, $\angle CBA$, and $\angle ACB$ intersect the circumcircle $\Omega$ at points $M$, $N$, and $K$, respectively.
- The segments $AB$ and $MK$ intersect at point $P$.
- The segments $AC$ and $MN$ intersect at point $Q$.

2. Use the properties of angle bisectors and the circumcircle:
- Since $M$ is on the circumcircle and is the intersection of the angle bisector of $\angle BAC$ with the circumcircle, $M$ is the midpoint of the arc $BC$ that does not contain $A$.
- Similarly, $N$ and $K$ are the midpoints of the arcs $AC$ and $AB$ that do not contain $B$ and $C$, respectively.

3. Apply the Angle Bisector Theorem:
- The Angle Bisector Theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides.
- Therefore, $BP : PA = MB : MA$ and $CQ : QA = MC : MA$.

4. **Use the fact that $M$, $N$, and $K$ are midpoints of the arcs:**
- Since $M$ is the midpoint of the arc $BC$, $MB = MC$.
- Similarly, $N$ and $K$ being midpoints of their respective arcs implies $NB = NA$ and $KA = KC$.

5. Establish the proportionality:
- From the Angle Bisector Theorem and the fact that $MB = MC$, we have:
$$BP : PA = MB : MA = MC : MA = CQ : QA$$
- This implies that $BP : PA = CQ : QA$.

6. **Conclude that $PQ \parallel BC$:**
- Since $BP : PA = CQ : QA$, by the converse of the Basic Proportionality Theorem (also known as Thales' theorem), the line segment $PQ$ must be parallel to $BC$.

$$\boxed{PQ \parallel BC}$$