Let O be the centre of the circle.
Then $\angle \mathrm{POR}={90}^o$ 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 $\mathrm{PO}=\mathrm{RO}=4\mathrm{cm}$. Let the length of $\mathrm{PR}$ be $x$ cm.
Then, by Pythagoras' Theorem, $x^2=4^2+4^2=2\times 4^2$ and so $x=4\sqrt{2}$.