Let $O$ be the centre of the circle.

Then $\angle POR=90^{\circ}$ as the angle subtended by an arc at the centre of a circle is twice the angle subtended by that arc at a point on the circumference of the circle.

So triangle $POR$ is an isosceles right-angled triangle with $PO=RO=4cm$. Let the length of $PR$ be $x$ cm.

Then, by Pythagoras' Theorem, $x^2=4^2+4^2=2 \times 4^2$ and so $x=4\sqrt{2}$.

*This problem is taken from the UKMT Mathematical Challenges.*