Since no sequence is given we need the help of options to identify the correct encoding.
h(⟨si…sn⟩)=Πni=1 Pi^{f}^{(si)} =P1^{f(s1)} * P2^{f(s2)} * .....*Pn^{f(sn)}
P1=2, P2=3, P3=5, P4=7 and so on...
g(a_{i})={3,5,7,9,11}.
Since in all the options there are first 3 prime nos. given, we can conclude that the sequence s goes up to n=3.
Now, f(s)=Πni=1Pi^{g(ai)}= P1^{g(s1)} * P2^{g(s2)} * .....*Pn^{g(sn)}. As it starts with P1 which is 2 and g(a_{i})≠0 for any case, so the value of f(s) can never be odd. Thus we can eliminate option A and C.
Next we see the powers of 2 in option B and D. Concentrating only on the power of 2 we find--->
B shows 8 which can be obtained if for i=1, g(a_{1})=3 and P1=2 we already know. 2^{3}=8.
D shows 10 which can be obtained if for i=1, g(a_{1})=1 and P1=2 (we know) , for i=2,g(a_{2})=0 and P2=3 (we know), for i=3, g(a_{3})=1 and P3=5 (we know). Then (2^{1})*(3^{0})*(5^{1})=10. But we know that g(a_{i})≠0 for any case. So D can't be the answer.
Hence B.