A maximum $AP \cdot PB$ cannot be attained if $AB$ intersects segment $O_1O_2$ because a larger value can be attained by making one of $A$ or $B$ diametrically opposite $P$ , which (as is easily checked) increases the value of both $AP$ and $PB$ . Thus, assume $AB$ does not intersect $O_1O_2$ .
Let $E$ and $F$ be the centers of the small and big circles, respectively, and $r$ and $R$ be their respective radii.
Let $M$ and $N$ be the feet of $E$ and $F$ to $AB$ , and $\alpha = \angle APE$ and $\epsilon = \angle BPF$
We have:
$$AP \times PB = 2r \cos{\alpha} \times 2R \cos{\epsilon} = 4 rR \cos{\alpha} \cos{\epsilon}$$
$AP\times PB$ is maximum when the product $\cos{\alpha} \cos{\epsilon}$ is a maximum.
We have $\cos{\alpha} \cos{\epsilon}= \frac{1}{2} [\cos(\alpha +\epsilon) + \cos(\alpha -\epsilon)]$
But $\alpha +\epsilon = 180^{\circ} - \angle EPF$ and is fixed, so is $\cos(\alpha +\epsilon)$ .
So its maximum depends on $cos(\alpha -\epsilon)$ which occurs when $\alpha=\epsilon$ . To draw the line $AB$ :
Draw a circle with center $P$ and radius $PE$ to cut the radius $PF$ at $H$ . Draw the line parallel to $EH$ passing through $P$ . This line meets the small and big circles at $A$ and $B$ , respectively.

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