smc

Consider the case where $P$ is on the hypotenuse of $AB$. Draw perpendicular lines from $P$ towards the sides. Using the Pythagorean Theorem, $$A{P}^2=a^2+a^2$$ $$B{P}^2=(x-a{)}^2+(x-a{)}^2$$ $$C{P}^2=a^2+(x-a{)}^2$$ This means $$s=4a^2-4ax+2x^2$$ $$2\cdot C{P}^2=4a^2-4ax+2x^2$$ Thus, $s=2\cdot C{P}^2$ when $P$ is on the hypotenuse of $AB$. Consider the case where $P$ is on the extension of $AB$. WLOG, let point $A$ be between point $P$ and point $B$. Extend $BC$ and draw perpendicular line from $P$. Also, draw point $D$, where $PD\parallel AC$ and $DA\parallel CB$. Using the Pythagorean Theorem again, $$A{P}^2=a^2+a^2$$ $$B{P}^2=(a+x{)}^2+(a+x{)}^2$$ $$C{P}^2=(a+x{)}^2+a^2$$ That means $$s=4a^2+4ax+2x^2$$ $$2\cdot C{P}^2=4a^2+4ax+2x^2$$ Thus, $s=2\cdot C{P}^2$ when $P$ is outside the hypotenuse. In summary, $A{P}^2+B{P}^2=2\cdot C{P}^2$, so the answer is $\boxed{s=2C{P}^2\text{ always}}$.