Problem 1571

1. Let $x = OT$. Since $OT > 1$, the smaller circle is outside the unit circle and touches it at point $T$.
2. Let $T'$ be the reflection of $T$ over the midpoint $Q$ of $AB$. Since $Q$ has the same power with respect to all three circles, we have:
$$QT^2 = QA \cdot QB = QP \cdot QO$$
This implies that the cross-ratio $(A, B; T', T) = -1$.
3. From the cross-ratio property, we have:
$$OT' \cdot OT = 1 \implies OT' = \frac{1}{x}$$
4. Since $Q$ is the midpoint of $TT'$, we get:
$$OQ = \frac{x + \frac{1}{x}}{2} = \frac{x^2 + 1}{2x}$$
5. Now, we remove $T'$ from the diagram as it was only needed to find $OQ$. Similarly, we have:
$$(A, B; P, Q) = -1 \implies OP \cdot OQ = 1 \implies OP = \frac{2x}{x^2 + 1}$$
6. Since $PA \cdot PB = (1 - PO)(1 + PO)$, the desired equality is equivalent to:
$$\left(1 - \frac{2x}{x^2 + 1}\right)\left(1 + \frac{2x}{x^2 + 1}\right) = \frac{\left(x - \frac{2x}{x^2 + 1}\right)^2}{x^2}$$
7. Simplifying the left-hand side:
$$\left(1 - \frac{2x}{x^2 + 1}\right)\left(1 + \frac{2x}{x^2 + 1}\right) = 1 - \left(\frac{2x}{x^2 + 1}\right)^2 = 1 - \frac{4x^2}{(x^2 + 1)^2}$$
8. Simplifying the right-hand side:
$$\frac{\left(x - \frac{2x}{x^2 + 1}\right)^2}{x^2} = \frac{\left(\frac{x(x^2 + 1) - 2x}{x^2 + 1}\right)^2}{x^2} = \frac{\left(\frac{x^3 - x + x - 2x}{x^2 + 1}\right)^2}{x^2} = \frac{\left(\frac{x^3 - 2x}{x^2 + 1}\right)^2}{x^2}$$
$$= \frac{\left(\frac{x(x^2 - 2)}{x^2 + 1}\right)^2}{x^2} = \frac{x^2(x^2 - 2)^2}{x^2(x^2 + 1)^2} = \frac{(x^2 - 2)^2}{(x^2 + 1)^2}$$
9. Equating both sides:
$$1 - \frac{4x^2}{(x^2 + 1)^2} = \frac{(x^2 - 2)^2}{(x^2 + 1)^2}$$
$$\left(\frac{x^2 - 1}{x^2 + 1}\right)^2 = \left(\frac{x^2 - 1}{x^2 + 1}\right)^2$$
This is true, thus proving the desired equality.

The final answer is $\boxed{|PA| \cdot |PB| = \frac{|PT|^2}{|OT|^2}}$.