Since you have no premises, but only a goal, let's focus on the goal and think about how you can get to something like that. Well, the main connective is a conditional ($\rightarrow$), so the strategy is probably to use $\rightarrow \: Intro$:
OK, so now your (new) goal is to get to $P$, using $(P \rightarrow Q) \rightarrow P$
Well, there is still very little you can do with $(P \rightarrow Q) \rightarrow P$ by itself: $\rightarrow Elim$ requires you to have the antecedent $P \rightarrow Q$, which you don't have. So, let's once again focus on the goal ($P$), and think about how we can prove something like that.
Now, here is a trick to remember: in order to prove an atomic statement like $P$, one strategy is to try and do a proof by contradiction. Here is how that works in Fitch:
OK, so now the (newest) goal is to prove $\bot$ using $(P \rightarrow Q) \rightarrow P$ and $\neg P$. OK, I won't show in detail how to do that, because at this point I want you to just look at the proof skeleton or proof plan that we have right now: do yourself a favor, and make sure to come up with such a proof plan for whatever proof you are working on! ... if your proof plan is good, the details will often pretty much fall out naturally.
For this particular proof, the rest isn't trivial, but it isn't hard either. Just think: "how do I get a $\bot$ given what I have? Well, that requires some statement and its negation. Hmm, I already have a negation $\neg P$. OK, so if I can get to $P$, I would be there. How can I get $P$? Oh, I have a conditional $(P \rightarrow Q) \rightarrow P$ whose consequent is $P$, so if only I could prove the antecedent $P \rightarrow Q$ I would be there. OK, how can I do that, presumably using $\neg P$?"
Well, I'll leave you to that. Good luck, and remember to rely on these proof skeletons!!