A problem L is NP-complete when:-
- L $\epsilon$ NP
- Every problem L' $\epsilon$ NP, L' is polynomial time reducible to L
For at least property 2 satisfied but do not know if 1st property (or not necessarily property 1) is satisfied by a problem L, then L $\epsilon$ NP-Hard.
For proving a problem satisfies property 1, best we can do is check if for any given input an algorithm can verify the solution in polynomial time, i.e., problem is polynomial time verifiable given a certificate of a solution as input.
For proving a problem satisfies property 2, it is difficult to prove all L' individually are polynomial- time reducible to L. Hence we make use of transitivity property of the polynomial reductions.
If we are able to show that a problem L'' already known to be in NP-complete to be polynomial-time reducible to the problem L, then L satisfies property 2 by transitivity.
Transitivity - If L1 is polynomial-time reducible to L2 and L2 is polynomial-time reducible to L3, then L1 is polynomial-time reducible to L3 by transitivity.
So, every L' is polynomial-time reducible to L'' (already NP-complete) and we prove L'' is polynomial-time reducible to L implies every L' is polynomial-time reducible to L.
In the case in the question, P (= L'') is already known to be in NP-complete. So, P is polynomial-time reducible to R(= L) implies R satisfies property 2. Hence, R $\epsilon$ NP-Hard for sure. For R $\epsilon$ NP-complete it requires to be NP too.