1. Define the polygons and their properties:
- Let $M$ be a bicentric polygon, meaning it has both an inscribed circle (incircle) and a circumscribed circle (circumcircle).
- Let $P$ be the polygon with vertices at the points of contact of the sides of $M$ with the incircle.
- Let $Q$ be the polygon formed by the external bisectors of the angles of $M$.

2. Establish the relationship between the polygons:
- Denote $M$ as $ABCD$, $P$ as $WXYZ$, and $Q$ as $EFGH$.
- Since $M$ is bicentric, the incircle touches the sides of $M$ at points $W, X, Y, Z$.
- The external angle bisectors of $M$ intersect the circumcircle at points $E, F, G, H$.

3. **Prove that $P$ and $Q$ are cyclic:**
- Since $M$ is bicentric, the points of tangency of the incircle with $M$ form a cyclic polygon $P$.
- The external angle bisectors of $M$ intersect the circumcircle at points $E, F, G, H$, forming a cyclic polygon $Q$.

4. Show parallelism and angle relationships:
- Consider the angle bisectors and perpendiculars from the incenter $I$ to the sides of $M$.
- Since $AI \perp WZ$, it follows that $WZ \parallel EH$.
- Similarly, $WX \parallel EF$.

5. Use cyclic quadrilaterals and angle chasing:
- Since $\angle HAI = \angle HDI = \angle GCI = 90^\circ$, quadrilaterals $HAID$ and $GCID$ are cyclic.
- For cyclic quadrilaterals, the opposite angles sum to $180^\circ$.
- Given that $ZDY$ is isosceles with bisector $DI$, we have $\angle AHI = \angle ADI = \angle IDC = \angle IGC$.

6. Establish collinearity and angle relationships:
- The equality of the first and last angles implies that $H, I, F$ and $G, I, E$ are collinear.
- Let $\angle AHI = \angle ADI = \angle IDC = \angle IGC = \phi$.
- Then $\angle ADC = 2\phi$ and $\angle ABC = 180^\circ - 2\phi$.

7. Prove similarity of triangles:
- Since $BXIW$ is cyclic, $\angle BWI = \angle BXI = 90^\circ$, and thus $\angle XIW = 2\phi$.
- Therefore, $\angle WYX = \phi$, implying $\angle WYX = \angle CGI \Rightarrow WY \parallel EG$.
- We conclude that $\Delta WXY \sim \Delta EFG$ and $\Delta WZY \sim \Delta EHG$.

8. Determine the center of homothety:
- Both similarities have the center of homothety at the intersection of $EW$ with $GY$.
- Since $WY$ and $EG$ are common sides of the respective similar triangles, $P$ and $Q$ are homothetic.

$$\boxed{\text{P and Q are homothetic.}}$$