I'm glad to see you gave MathJax a try :D
"Determine [a set]" means to characterize the elements of the set so you can "easily" find out if a given explicit element is in the set or not. It is kind of a vague exercise formulation, basically a "Prove that $A=B$" without giving you $B$, but letting you find a suitable $B$ yourself. So basically, there is no "the right answer", only "the answer the author had in mind" and it is basically your goal to get as close as possible.
Example: When you are given $(x_1,y_1)-(x_2,y_2):=(x_1-x_2,y_1-y_2)$, $||(x,y)||:=x^2+y^2$ and $$U_r((x,y)):=\{(x_1,y_1):||(x_1,y_1)-(x,y)||\leq r^2\}$$
for $x,x_1,x_2,y,y_1,y_2,r\in\mathbb{R}$, and I say "Determine $U_r((x,y))$.", then a valid answer would be
$$U_r((x,y))=\{(x_1,y_1):(x_1-x)^2+(y_1-y)^2\leq r^2\}$$
Bonus points if you notice that $U_0((x,y))=\{(x,y)\}$ and $U_r((x,y))=\{\}$ if $r<0$. (For $r>0$ $U_r((x,y))$ describes the disk with center $(x,y)$ and radius $r$ in $\mathbb{R}^2$.
So in your case, you are basically expected to use the definitions again and again until there is nothing more to do. Any hint or solution would be in your book as well, so I'm leaving it out here.
EDIT: Oh wait, we have spoilers, so here is what I would do :D
First let me note that I believe your first step is wrong.
$$P(f)(P(\mathbb{R}))=\{P(f)(A):A\in P(\mathbb R)\}$$
This is important: Since $P(\mathbb{R})\notin P(\mathbb{R})$ but $P(\mathbb{R})\subseteq P(\mathbb{R})$, you don't get to use the definition of $P(f)$ but the definition of $f(A)$. Of course it could be that the author made some remark about notation, so that your first step is actually right an I am wrong. You can only find that out by looking at the solution later.
Rest is spoiling.
Since $P(f)(A)= f(A) = \{x^2:x\in A\}$ we have
$$P(f)(P(\mathbb{R}))=\{\{x^2:x\in A\}:A\in P(\mathbb{R})\}$$
Now any nonnegative number $y$ has a nonnegative number $x$ with $y=x^2$ ($x=\sqrt{y}$), so for each $B\in P(\mathbb{R}_{\geq 0})$ there is a $A\in P(\mathbb{R})$ with $B=\{x^2:x\in A\}$. On the other hand, since $x^2$ is never negative, there are no other kinds of sets in $\{\{x^2:x\in A\}:A\in P(\mathbb{R})\}$. We conclude
$$P(f)(P(\mathbb{R}))=\{B:B\in P(\mathbb{R}_{\geq 0})\}=P(\mathbb{R}_{\geq 0})$$
