1. Let $AB$ be the diameter of the circle with center $O$ and radius $1$. Since $AB$ is the diameter, $\angle AOB = 180^\circ$.
2. Let $C$ and $D$ be points on the circle such that $AC$ and $BD$ intersect at point $Q$ inside the circle, and $\angle AQB = 2 \angle COD$.
3. Let $\angle COD = x$. Then, $\angle AQB = 2x$.
4. Since $AC$ and $BD$ intersect at $Q$, we can use the fact that the sum of the angles around point $Q$ is $360^\circ$. Therefore, $\angle CAQ + \angle QAB + \angle QBD + \angle DQC = 360^\circ$.
5. Let $\angle CAO = \angle ACO = m$ and $\angle DBO = \angle BDO = n$. Since $\angle CAO$ and $\angle ACO$ are angles subtended by the same arc $CO$, they are equal. Similarly, $\angle DBO$ and $\angle BDO$ are equal.
6. In $\triangle AQB$, the sum of the angles is $180^\circ$. Therefore, $m + n = 180^\circ - 2x$.
7. Since $\angle COD = x$, the angles around point $O$ must add up to $360^\circ$. Therefore, $2m + 2n = 360^\circ - 2x$.
8. Simplifying, we get $m + n = 180^\circ - x$.
9. From steps 6 and 8, we have two equations:
$$m + n = 180^\circ - 2x$$
$$m + n = x$$
10. Equating the two expressions for $m + n$, we get:
$$180^\circ - 2x = x$$
11. Solving for $x$, we get:
$$180^\circ = 3x \implies x = 60^\circ$$
12. Therefore, $\angle COD = 60^\circ$ and $\angle AQB = 2 \times 60^\circ = 120^\circ$.
13. Since $\angle COD = 60^\circ$, triangles $POD$ and $COP$ are $30^\circ-60^\circ-90^\circ$ triangles with $DO = CO = 1$ (radius of the circle).
14. In a $30^\circ-60^\circ-90^\circ$ triangle, the ratio of the sides opposite the $30^\circ$, $60^\circ$, and $90^\circ$ angles are $1 : \sqrt{3} : 2$.
15. Therefore, the length of $PO$ (the side opposite the $60^\circ$ angle) is:
$$PO = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$

The final answer is $\boxed{\frac{2\sqrt{3}}{3}}$.