You are given the premise (P > Q) > R. Since you want to show a conditional, P > (Q > R), assume as you attempted the antecedent, P. Then you need to derive the consequent, Q > R.
However, that consequent is also a conditional. So start another subproof within the first subproof by assuming the antecedent Q. Now the goal is to derive R.
At this point you have two subproofs one inside the other and you need to derive R. If you had P > Q, you could derive R from the premise. So, first you have to derive P > Q.
Since that is a conditional make a third subproof inside the second one by assuming P. You already have Q as an assumption so you can use reiteration to derive it. That gives you P > Q. From there you can derive R and from that result Q > R. From that the final goal P > (Q > R) can be derived.
Here is a proof using a Fitch-style proof checker:

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/