Solution:

The condition $P(n) = \sqrt{n}$ means $n$ is the square of a prime $q,$ and likewise $n + 48 = p^2$ for a prime $p.$ Then $48 = p^2 - q^2 = (p - q)(p + q).$ Checking the same-parity factorizations of $48,$ only $p - q = 2,\ p + q = 24$ yields primes, giving $(p, q) = (13, 11)$ and $n = 121.$ So there is exactly one such $n.$

Thus, the correct answer is B.