1. Set up the circle using complex numbers

Let the given circle have center at the origin and radius $2$.

Represent the vertices of the regular octagon as $A_k=2\omega^k,\qquad k=0,1,2,\dots,7$ where $\omega=e^{2\pi i/8}=e^{i\pi/4}.$

Let the moving point $P$ on the same circle be $P=2z,$ where $|z|=1$.

Then the distances are $PA_k=|2z-2\omega^k|=2|z-\omega^k|.$ So the required product is $\prod_{k=0}^7 PA_k = 2^8 \prod_{k=0}^7 |z-\omega^k|.$

2. Use the polynomial whose roots are the 8th roots of unity

Since $\omega^0,\omega^1,\dots,\omega^7$ are the roots of $x^8-1=0,$ we have $\prod_{k=0}^7 (z-\omega^k)=z^8-1.$ Taking modulus, $\prod_{k=0}^7 |z-\omega^k|=|z^8-1|.$ Therefore, $\prod_{k=0}^7 PA_k = 2^8 |z^8-1|.$

3. Maximize $|z^8-1|$ when $|z|=1$

Write $z=e^{i\theta}.$ Then $z^8=e^{i8\theta},$ so $|z^8-1|$ is the distance between the point $e^{i8\theta}$ on the unit circle and the point $1$.

This distance is at most $2$, with equality when $e^{i8\theta}=-1.$ Hence, $\max |z^8-1|=2.$

Therefore, $\max \prod_{k=0}^7 PA_k = 2^8 \cdot 2 = 2^9 = 512.$

4. Final answer

The maximum value of $PA_1\cdot PA_2\cdots PA_8$ is $\boxed{512}.$