1. Representation of Vertices on the Unit Circle:
Let $a, b, c$ be complex numbers representing the vertices $A, B, C$ of the triangle $\triangle ABC$ on the unit circle centered at the origin $0$. This means $|a| = |b| = |c| = 1$.

2. Similarity of Triangles:
Given that $\triangle PAB \sim \triangle PCA$ in the same orientation, we can write the similarity condition as:
$$\frac{p-a}{b-a} = \frac{p-c}{c-a}$$
Solving for $p$, we get:
$$\frac{p-a}{b-a} = \frac{p-c}{c-a} \implies (p-a)(c-a) = (p-c)(b-a)$$
Expanding and simplifying:
$$pc - pa - ac + a^2 = pb - pa - bc + ac$$
$$pc - bc = pb - pa + a^2 - ac$$
$$p(c - b) = a^2 - ac + bc$$
$$p = \frac{a^2 - ac + bc}{c - b}$$
Simplifying further:
$$p = \frac{bc - a^2}{b + c - 2a}$$

3. **Expression for $\frac{p-a}{p-0}$:**
We need to find the expression for $\frac{p-a}{p}$:
$$\frac{p-a}{p} = \frac{\frac{bc - a^2}{b + c - 2a} - a}{\frac{bc - a^2}{b + c - 2a}}$$
Simplifying the numerator:
$$\frac{bc - a^2 - a(b + c - 2a)}{bc - a^2} = \frac{bc - a^2 - ab - ac + 2a^2}{bc - a^2} = \frac{bc - ab - ac + a^2}{bc - a^2}$$
$$\frac{p-a}{p} = \frac{(a-b)(a-c)}{bc - a^2}$$

4. Conjugate and Orthogonality:
Since $a, b, c$ lie on the unit circle, their conjugates are their reciprocals. We need to show that:
$$\frac{p-a}{p} = -\overline{\left(\frac{p-a}{p}\right)}$$
This implies:
$$\frac{(a-b)(a-c)}{bc - a^2} = -\frac{\overline{(a-b)(a-c)}}{\overline{bc - a^2}}$$
Since $a, b, c$ are on the unit circle, we have:
$$\overline{a} = \frac{1}{a}, \quad \overline{b} = \frac{1}{b}, \quad \overline{c} = \frac{1}{c}$$
Therefore:
$$\overline{(a-b)(a-c)} = \left(\frac{1}{a} - \frac{1}{b}\right)\left(\frac{1}{a} - \frac{1}{c}\right) = \frac{(b-a)(c-a)}{abc}$$
And:
$$\overline{bc - a^2} = \frac{1}{bc} - \frac{1}{a^2} = \frac{a^2 - bc}{a^2bc}$$
Thus:
$$-\frac{\overline{(a-b)(a-c)}}{\overline{bc - a^2}} = -\frac{\frac{(b-a)(c-a)}{abc}}{\frac{a^2 - bc}{a^2bc}} = -\frac{(b-a)(c-a)}{a^2 - bc}$$
This shows that:
$$\frac{(a-b)(a-c)}{bc - a^2} = -\frac{(b-a)(c-a)}{a^2 - bc}$$
Hence:
$$\frac{p-a}{p} = -\overline{\left(\frac{p-a}{p}\right)}$$
This implies that $\angle APO = 90^\circ$.

$\blacksquare$