Within RP, P generates a maximal ideal. Since localization and product commute, the image of Pn in RP is the image of P, raised to the nth power. The power of a maximal ideal is primary, so the image of Pn is primary beneath the image of P.
The contraction of a primary ideal is primary, so P[n] is primary in R. Since P is the radical of Pn, which is contained in P[n], P[n] is P primary.
Assume Pn has a primary decomposition H1 H2 H3 etc. Take radicals, and P is the intersection of the primes lying over our primary ideals. Thus P contains some other prime ideal Q, which contains Pn. However, P is the radical of Pn, hence Q = P. At least one of the primary ideals is P primary.
Since P is the intersection of the prime ideals, all the other primes contain P. Therefore P is an isolated prime. Another way to see this is to note that P is a minimal prime above Pn, being the radical of Pn, and a minimal prime is isolated.
Use second uniqueness to describe the primary ideal below P. Saturate Pn to get P[n]. This is one of the primary ideals in the decomposition of Pn, and it is P primary. In fact it is the smallest primary ideal (containing Pn) below P. If Pn is itself P primary it must equal P[n]. Conversely, if Pn = P[n], then Pn is P primary, creating a trivial primary decomposition.
Let H = P[m+n]. Since the localization of Pm+n is primary below a maximal ideal, H is primary below P.
Both H and J have the same image in RP, namely the image of Pm+n. Since H is the saturation, it contains J. Don't assume H = J, because different ideals can lead to the same image in RP.
Ok, how do we get to H? J is the intersection of primary ideals, so take radicals across the board. Remember that radical and intersection commute, so the right side is the intersection of primes lying over our primary ideals. The left side takes the radical of a product of two ideals P[m] and P[n]. This is equal to the radical of their intersection. Assume m is smaller, whence P[m] wholly contains P[n]. The latter becomes the intersection of the two ideals, and its radical is P. Thus an intersection of prime ideals equals P, and one of them, say Q, is contained in P. As before, P defines the radical of J, and there can be no prime ideals stricly between P and J, hence Q = P. One of the primary ideals is P primary.
As before, all the other primes, lying over primary ideals, lying over J, must contain P, hence P is isolated. Apply second uniqueness and H is the P primary ideal that participates in the decomposition of J. Once again H = J iff J is P primary.