The answer is $2\sqrt{2}$. Rotating $OPQR$ clockwise for $90^{\circ}$ around $R$ gets the square $RSTO$. By symmetry, we find $CR=BP=\sqrt{2}$.

Now let's show that $CS=1$ and $CS\perp SQ$. (These two conclusions are essentially equivalent). There are two ways. The first is to apply law of cosines on $\triangle{CQR}$: $$\cos\angle{QRC}=\frac{1^2 + (\sqrt{2})^2-(\sqrt{5})^2}{2\times 1\times \sqrt{2}}=-\frac{\sqrt{2}}{2}$$ Therefore $$\angle{QRC}=135^\circ\implies\angle{SRC}=45^\circ\implies CS=1$$ Another way is to use Apollonius’ Theorem (see # 3866): $$SC^2 + (\sqrt{5})^2=2\times(1^2 + (\sqrt{2})^2)\implies SC = 1$$