The slope of $PP\text{'}=-1$    ( As, $PP\text{'}$ is perpendicular to the given line)

$\Rightarrow \frac{k-5}{h-3}=-1 \Rightarrow k+h=8 ...\left(1\right)$

Mid-point of $P{P}^{\text{'}}$ lies on the given line,

$\Rightarrow \left(\frac{h+3}{2}\right)-\left(\frac{k+5}{2}\right)+1=0 \Rightarrow h+3-k-5+2=0 \Rightarrow h-k=0 ...\left(2\right)$Solving equation $\left(1\right)$ and $\left(2\right)$ we get, $h=4, k=4$

Which satisfy the equation ${\left(x-2\right)}^2+{\left(y-4\right)}^2=4$.