Solve SUBSET SUM for Reciprocals of Primes
The observation is that the denominator of the reduced fraction $1/p_{i_1} + \cdots + 1/p_{i_m}$ is $p_{i_1} \cdots p_{i_m}$. To see this, it suffices to notice that the (unreduced) numerator isn't divisible by any $p_{i_j}$. Indeed, the numerator is simply
$$
\frac{p_{i_1} \cdots p_{i_m}}{p_{i_1}} + \cdots + \frac{p_{i_1} \cdots p_{i_m}}{p_{i_m}}.
$$
All terms but the $j$th are products of $p_{i_j}$, but the $j$th term isn't, so the numerator isn't divisible by $p_{i_j}$.

As a consequence, if $\alpha/P$ is a sum of reciprocals of distinct primes, then we can read off the primes from the prime factorization of the reduced form of $\alpha/P$. We can compute this reduced form efficiently using GCD. Since we know $p_1,\ldots,p_n$, we can figure out whether the denominator is a product of a subset of $\{p_1,\ldots,p_n\}$, and if so, it only remains to check whether $\alpha/P$ is equal to the sum of reciprocals of prime in this subset.