$P(a,b)$ means put $x=a,y=b$ define $f(0)=c$ $P(0,0) \Rightarrow 2f(c)=c $ , $P(0,f(c)) \Rightarrow f(c)=0 \Rightarrow c=0$ (because $c-f(c)=f(c)$) $P(x,0) \Rightarrow f(f(x))=0 \Rightarrow f(f(x)-y)=f(y)$ now put $x=0$ so $f(y)=f(-y)$ so $f(y-f(x))=f(y)$ and then if $\exists a \mid f(a) \neq 0$ then $f$ is periodic!!

And all such functions can be described. Either $f \equiv 0$, or there exists some positive integer $t> 1$ such that $f(0)=0$, $f(t)=0$, $f(k) = n_kt$ for $1\leq k \leq t-1$ and some integer $n_k$'s, and $f$ is periodic of period $t$.

let $y=0$,$f(f(x))=\frac{f(0)}{2}$ let $f(0)=c,x=c,y=0$then $f(\frac{c}{2})=\frac{c}{2}$ let $x=y=\frac{c}{2}$ then$f(0)=f(\frac{c}{2})-f(f(\frac{c}{2}))=0$ hence $f(f(x))=0$ and we get $f(f(x)-y)=f(y)$ let $x=0$,then$f(-y)=f(y)$ f is odd let us suppose there exists $c\ne 0,f(x)=c$ then $f(y)=f(c-y)=f(y-c)$ hence f is a periodic function with period c,proved.

$P(x,y)=f(f(x-y)=f(y)-f(f(x))$. Now put $x=y=0$ $P(0,0):2f(f(0))=f(0)-f(f(0))$ assume that $f(0)=a$ this implies that $2f(a)=a$. clearly $f(a)=0$ Now use it and prove that this function is periodic.

$P(x,f(x))\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(f(x))=0$ $P(0,x)\Rightarrow f(x)=f(-x)$ $P(x,y)\Rightarrow f(y-f(x))=f(y)$ So either $f(x)=0$, which is bounded, or $f$ is periodic, meaning it is also bounded.