To get the list of the first 100 primes, we can use the Prime function and assign it to a variable primes:
In[12]:= primes = Prime[Range[100]]
Out[12]= {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, \
59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, \
131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, \
199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, \
281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, \
373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, \
457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541}
Evaluating the given expression in Mathematica using Sum and Exp functions, with N to get the numerical value:
In[16]:= N[Exp[Sum[1/(k p^k), {k, 1, 100}, {p, primes}]]]
Out[16]= 11.2676
I don't know what is s in your equation, so I couldn't insert it in the above expression (Sorry!).
For the upper bound of the k sum (over reals), I have taken 100 and not infinity, since the series seems to be converged (up to four decimal places) for n=100.
One can check the result for the sum over k for three different upper limits, n=10, 100, 1000:
In[18]:= Table[N[Exp[Sum[1/(k p^k), {k, 1, 10^n}, {p, Prime[Range[100]]}]]], {n, {1,2,3}}]
Out[18]= {11.2667, 11.2676, 11.2676}
