The first few primes are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (P<50).
The sum of consecutive primes that are themselves are primes can be identified:
p1=1; p2=1+2=3; p4=1+2+3+5=11; p6=1+2+3+5+7+11=29; p8=1+2+3+5+7+11+13+17=59; p10=1+2+3+...+17+19+23+101, ...
My understanding of the questions is that P1={ p1 }; P2={ p1 p2 }; P3={ p1 p2 p4 }; P4={ p1 p2 p4 p6 }; P5={ p1 p2 p4 p6 p8 }; P6={ p1 p2 p4 p6 p8 p10 }. P2 and P4 are defined in this list. In Pn (P sub n), n defines the number of elements, and since n is a finite number (natural number belonging to N), the sets are finite by definition. In the sums p sub k it follows that k will be even (apart from p1) because, after 2, all primes are odd, adding another prime will automatically result in an even number, so a further prime must be added to get an odd number which may be prime. However, not all odd numbers obtained this way will necessarily be prime because not all odd numbers are prime.
What I'm not sure of is whether P3={ p1 p2 p4 } should in fact be defined as P4 because the highest pk is p4. If this is indeed the case then the list given earlier should define the sets for P1, P2, P4, P6, P8. P10 instead of P1, P2, P3,... Under this assumption, P4={ p1 p2 p4} while P2 is unaffected. However, part 4 suggests that P3 and P5 are definable. This takes us back to the original definition. P3 has only three elements under this definition: p1, p2, p4. If p1, p2 and p4 are themselves considered as sets of the primes comprising the sums, then, assuming duplicates are allowed, we have P3={ {1 } { 1 2 } { 1 2 3 5 } }. Five elements from these are, for example, { 1 1 1 2 2 } which also applies to P5.