Let $f:R\to R$ be a function such that for some $\text{p}>0,f(x+p)=\frac{f\left(x\right)-5}{f\left(x\right)-3}$ for all $x\in R.$ Then, period of f is

3 does not belongs to the range of $f\Rightarrow 2$ also does not belongs to the range of f for  otherwise $f\left(x\right)$ for some $x\in R.$ If $f(x+2p)=f\left(x\right),$  then

$f(x+2p)=f(x+p+p)=\frac{\frac{f\left(x\right)-5}{f\left(x\right)-3}-5}{\frac{f\left(x\right)-5}{f\left(x\right)-3}-3}=\frac{-4f\left(x\right)+10}{-2f\left(x\right)+4}=\frac{2f\left(x\right)-5}{f\left(x\right)-2}$

So that ${\left(f\right(x)-2)}^2=-1$ which is absured

Hence $f(x+2p)\ne f\left(x\right)$ Similarly $f(x+3p)\ne f\left(x\right)$

$f(x+4p)=\frac{f(x+3p)-5}{f(x+3p)-3}=\frac{(\frac{3f\left(x\right)-5}{f\left(x\right)-1}-5)}{(\frac{3f\left(x\right)-5}{f\left(x\right)-1}-3)}=\frac{-2f\left(x\right)}{-2}=f\left(x\right)$