The part (a) and part (b) are similar by considering the Fixed-Point Iteration of $P(x)=x.$ Note that $ran(P)=\mathbb{R}$ and for any $\mathbb{N}\ni y_1\in Ran(f)$ there must exist some $y_2\in Ran(f)$ such that $P(y_2)=y_1.$ Hence we can obtain a (not unique) sequence of $y_n$ satisfying $$P(y_{i+1})=y_i,i=1,2,\cdots.$$ (a): Now we suppose that $|ran(f)|=+\infty$ and select $y_1\gg 1$ such that $P(x)=y_1$ has only one real root. Then we obtain a sequence of $y_i,i=1,2\cdots$ satisfying $$0\leq y_i< y_1, \forall i=1,2,\cdots.$$By Pigeonhole Principle, there must exist $j>i$ such that $y_j=y_i,$ which implies $y_{j-i+1}=P^{i-1}(y_j)=P^{i-1}(y_i)=y_1.$ This is a contradiction with $y_{j-i+1}<y_1.$ (b):If $P(x)=x$ has only one real root and $f$ is not a constant, then the sequence is strictly monotonic. This is a contradiction with the finiteness of $ran(f).$ (c):$n$ is odd: For any fixed $a_1<a_2<\cdots<a_n\in \mathbb{N}.$ Let $$P(x)-x=(x-a_1)(x-a_2)\cdots(x-a_n).$$Denote: $$f(x)=a_i, \forall ~~~x\in \left\{x\in \mathbb{R} | P(x)=a_i\right\},i=1,2,\cdots n.;$$$$f(x)=a_1, ~~~otherwise.$$ $n$ is even: For any fixed $a_1<a_2<\cdots<a_n\in \mathbb{N}.$ Let $$P(x)-x=(x-a_1)(x-a_2)\cdots(x-a_n)^2.$$Denote: $$f(x)=a_i, \forall ~~~x\in \left\{x\in \mathbb{R} | P(x)=a_i\right\},i=1,2,\cdots n. ;$$$$f(x)=a_1, ~~~otherwise,$$