Solution:

a.
Extend line $A P$ to some point $Q$ on the opposite side of $P$ from $A$. Let $R$ be the intersection of lines $Q O$ and $P B$, and let $S$ be the intersection of lines $A R$ and $Q B$, as shown below.

By Ceva's theorem,
$$\frac{Q P}{P A} \cdot \frac{A O}{O B} \cdot \frac{B S}{S Q} = 1$$
It follows that
$$\frac{Q P}{P A} = \frac{Q S}{Q B}$$
which means $\triangle Q P S \sim \triangle Q A B$, therefore $\angle Q P S = \angle Q A B$ and so $P S \parallel A B$.

b.
Let $P A$ and $P B$ meet the circle again at points $Q$ and $R$, respectively, and let $S$ be the intersection of $A R$ and $B Q$.

Then $\angle A Q B = \angle A R B = 90^\circ$ since both angles are inscribed in a semicircle, so $S$ is the orthocenter of $\triangle A B P$, which means $P S$ is the desired perpendicular.