$P(a,a,a,a)$ gives $f(0)=0$. If there's a $x$ such that $f(x)<0$. $P(a,a,c,d)$ gives $f(a-d)f(a-c)<=(a-c)f(a-d)$ Now if we take $d$ such that $a-d=x$ we have $f(a-c)<=a-c$ Of we take $d$ such that $f(a-d)>0$ we have $f(a-c)>=a-c$ So $f(x)=x$ If there is no $x$ such that $f(x)<0$ then we have $P(a,b,a,b)$ gives $f(a-b)^2+f(a-d)f(b-a)<=0$ Which gives $f(a-b)=0$

We claim the only solution is given by $f(x)=x$ or $f(x)=0$. Clearly these work. Assume $f$ is not identically $0$. If we plug in $a=c$, we get \[f(a-d)\left[f(a-b)+f(b-a)\right]\leq 0\]Letting $y=a-d$ and $x=b-a$ (and noting these are independent), we get \[f(y)(f(x)+f(-x))\leq 0\]Assuming that $f(x)\neq -f(-x)$, we get that $f(y)$ must have a constant sign as $y$ ranges over $\mathbb R$. If we take $y$ such that $f(y)\neq 0$, However, this means that $f(x)+f(-x)$ shares that sign as well, a contradiction. Thus, $f(x)=-f(-x)$. Now, we plug in $a=b$. Then, we get \[f(a-d)f(a-c)\leq(a-c)f(a-d)\]Thus, letting $x=a-d$ and $y=a-c$, we get \[f(x)\left[f(y)-y\right]\leq 0\]However, we also know that \[0\leq -f(x)=\left[f(y)-y\right]=f(-x)\left[f(y)-y\right]\leq 0\]which implies everything is an equality. As $f$ is not identically $0$, we get $f(y)=y$ for all $y\in\mathbb R$, as desired.

I think if we make a=0 and this problem is gonna to be pretty easy