a.
We denote the angles of $\triangle ABC$ by $\angle A$, $\angle B$, $\angle C$. Since $AI$ is a diameter, we have $\angle ACI = \angle ABI = 90^\circ$. Also, $\angle AIC = \angle B$ (both standing on arc $AC$), and $\angle CAI = \angle CBI$ (both standing on arc $CI$). Therefore,

$\triangle ACI \sim \triangle ADB \sim \triangle BQI$. Similarly, $\triangle ABI \sim \triangle ADC \sim \triangle CQI$ and
$$\frac{|AC|}{|CI|} = \frac{|AD|}{|DB|} = \frac{|BQ|}{|QI|} \quad \text{as well as} \quad \frac{|AB|}{|BI|} = \frac{|AD|}{|DC|} = \frac{|CQ|}{|QI|}$$
which gives $|AD| \cdot |QI| = |CD| \cdot |CQ| = |BD| \cdot |BQ|$.

b.
Solution 1:
Let $X$ denote the intersection point of $AI$ and $BC$. Then, by power of $X$ with respect to the circumcircle of $ABC$ (or $\triangle AXB \sim \triangle CXI$) we have $|AX| \cdot |XI| = |CX| \cdot |XB|$. By power of $X$ with respect to the circumcircle of $CBO$, we have $|OX| \cdot |XP| = |CX| \cdot |XB|$. Combining both equations, we obtain
$$|OX| \cdot |XP| = |AX| \cdot |XI| \quad \text{i.e.} \quad \frac{|AX|}{|OX|} = \frac{|XP|}{|XI|}$$

By the Intercept Theorem, or $\triangle OMX \sim \triangle ADX$, we have
$$\frac{|AX|}{|OX|} = \frac{|DX|}{|MX|}, \quad \text{hence} \quad \frac{|XP|}{|XI|} = \frac{|DX|}{|MX|}$$
This implies similarity of triangles $MXI$ and $DXP$, hence $MI \parallel DP$.

Solution 2:
Substituting $|CD| = |BC| - |BD|$ and $|BQ| = |BC| - |CQ|$ in the equation $|CD| \cdot |CQ| = |BD| \cdot |BQ|$ from part (a) yields $|BD| = |CQ|$. Since $M$ is the midpoint of $BC$, it is then also the midpoint of $DQ$. We define $J$ on the extended line $QI$ such that $I$ is the midpoint of segment $QJ$. Then $MI \parallel DJ$ (mid-line in $\triangle DQJ$). It remains to prove that $J$ is on the line $DP$.

We first prove that $I$ is the incentre of $\triangle CPB$. Indeed, we have
$\angle CBP = \angle COP$ and $\angle BCP = \angle BOP$ in $\omega$
$\angle COP = 2\angle CBI$ and $\angle BOP = 2\angle BCI$ in $\Gamma$
so $\angle CBP = 2\angle CBI$ and $\angle BCP = 2\angle BCI$ hence $BI$ and $CI$ are angle bisectors and $I$ is the incentre of $\triangle CPB$.
Because $AI$ is a diameter of $\Gamma$, $AC \perp CI$ and $AB \perp BI$. Therefore, $AC$ and $AB$ are external angle bisectors of $\triangle CPB$ and so $A$ is an excentre.

Consider the inradius and exradius given by $IN$ and $AL$, both perpendicular to $PC$. Then triangles $\triangle PNI$ and $\triangle PLA$ are similar, which implies
$$\frac{|PI|}{|PA|} = \frac{|IN|}{|AL|} = \frac{|IQ|}{|AD|} = \frac{|IJ|}{|AD|}.$$
Because $IJ$ is parallel to $AD$, we have $\angle PIJ = \angle PAD$. Hence, triangles $\triangle PIJ$ and $\triangle PAD$ are similar and $P$, $J$, $D$ are collinear.

Solution 3:
From (a) we get $|BD| = |CQ|$, see Solution 2. This implies $|BQ| = |CD|$ and that $M$ is the midpoint of $QD$. We extend the line $QI$ to intersect $PD$ at $J$. To prove $MI \parallel PD$, it is sufficient to show $|QI| = |IJ|$. This is equivalent to $\frac{|QI|}{|AD|} = \frac{|PI|}{|PA|}$, because $\frac{|IJ|}{|AD|} = \frac{|PI|}{|PA|}$ follows from $QJ \parallel AD$.

Using (a), we obtain
$$\frac{|QI|}{|AD|} = \frac{|BD| \cdot |BQ|}{|AD|^2} = \frac{|BD| \cdot |CD|}{|AD|^2} = \cot B \cot C.$$

On the other hand,
$$\frac{|PI|}{|PA|} = \frac{[CPI]}{[CPA]} = \frac{\frac{1}{2}|CP| \cdot |CI| \sin \angle PCI}{\frac{1}{2}|CP| \cdot |CA| \sin \angle PCA} = \frac{|CI| \sin \angle PCI}{|CA| \sin \angle PCA},$$
and $\frac{|CI|}{|CA|} = \cot \angle CIA = \cot B$, while
$$\begin{aligned} \angle PCA &= 180^\circ - \angle CAO - \angle CPO = 180^\circ - \angle ACO - \angle CBO \\ &= 180^\circ - \angle ACO - \angle OCB = 180^\circ - \angle C \end{aligned}$$
and so $\angle PCI = \angle PCA - 90^\circ = 90^\circ - \angle C$. Hence
$$\frac{\sin \angle PCI}{\sin \angle PCA} = \frac{\sin (90^\circ - \angle C)}{\sin (180^\circ - \angle C)} = \frac{\cos C}{\sin C} = \cot C.$$
Therefore,
$$\frac{|PI|}{|PA|} = \frac{|CI|}{|CA|} \cdot \frac{\sin \angle PCI}{\sin \angle PCA} = \cot B \cot C = \frac{|QI|}{|AD|},$$