$a)$$P(x,x)$ gives $f(x-f(x))=f(0)$. $P(0,x-f(x))$ gives $f(-f(0))=f(f(x)-x)$. $P(0,0)$ gives $f(-f(0))=f(0)$. Thus we have $f(f(x)-x)=f(0)$. Plugging $x=0$ gives $f(0)=f(f(0))$. $P(x,f(0))$ gives $c(f(x)-f(0))=0$. If there exists $x$ satisfying $f(x)\neq f(0)$, we have $c=0$. Otherwise $f(x)$ is constant. Therefore $c=0$. $b)$$P(x+y,y)$ gives $f(x+y-f(y))=f(x)$. If $T$ is the period, we have $y-f(y)=Tg(y)$ where g is a function from Z to Z. Putting $f(x)=x-Tg(x)$ in the equation gives $2y=T(g(y)+g(x+y-g(y)T)-g(x-y))$. Because $g$ is Z to Z, we have $2y=h(x,y)T$ where $h(x,y)=g(y)+g(x+y-g(y)T)-g(x-y)$ is a function from ZxZ to Z. Obviously, this cannot be true for all $y$. For example, $0<y<T/2$. Thus $f$ cannot be periodic. NOTE: Solution of b is wrong. Because $2y=T(g(y)+g(x+y-g(y)T)-g(x-y))$ is wrong. Can anyone check this solution?

We claim that the only solution is $c=0$. See that if $c=0$, then $f(x)=x$ works. Let $c\neq 0$ and let $P(x,y)$ be the assertion. Lemma: Let $a$ and $b$ be real numbers such that $f(a)=f(b)$. Then $f(x-a)=f(x-b)$ for each real number $x$. Proof: $P(x,a)$ and $P(x,b)$ directly gives the desired result. $P(1,1)$ gives $f(1-f(1))=f(0)$. By Lemma we can say that $f(x-1+f(1))=f(x)$. Replace $x$ with $x-f(1)$. Then $f(x-1)=f(x-f(1))$. $P(x,1)$ gives $f(x-f(1))=f(x-1)+c(f(x)-f(1))\Rightarrow c(f(x)-f(1))=0$. We assumed $c\neq 0$. Thus, $f(x)=f(1)$ for all real numbers. Thus, $f$ is constant. Contradiction. Hence $c=0$ is the only valid solution.

See that $f(x)=\{x\}$ works when $c=0$. Thus the function can be periodic.

For part a, we claim that the only possible value of $c$ is $0$, this obviously works. Let $P(x,y)$ be the assertion. $P(x,0)$, $P(x,x)$ and $P(x,x-f(x))$ give us $f(f(x))=f(x)$. Now, $P(x,y)$ and $P(x,f(y))$ imply $f(x-y)=f(x-f(y))$ and therefore $c(f(x)-f(y))=0$ for all $x,y\in\mathbb{R}$. Since $f$ is not a constant function, thus $c=0$. For part b, one could use $f(x)=x-\lfloor x\rfloor \forall x\in\mathbb{R}$.