The following solution is based on the idea shared by @Sachin Mittal sir.
Given,
$g_y(z) = (1 – \beta + \beta z)^N$
$= g_y(z) = ((1 – \beta) + \beta z)^N$
Using binomial expansion,
$= g_y(z) = \sum_{k=0}^{N}{\binom{N}{k}(1 – \beta)^{N-k}(\beta z)^k}$
$= g_y(z) = \sum_{k=0}^{N}{\binom{N}{k}(1 – \beta)^{N-k}(\beta)^k(z)^k}$
Let $a = (1 – \beta)$ and $b = \beta$
$= g_y(z) = \sum_{k=0}^{N}{\binom{N}{k}a^{N-k}b^k(z)^k}$
comapring $g_x(z) = \sum_{j=0}^{N}p_jz^j$ to the above $eq^n$ we know that each of the terms in the above summation will give $P(Y = j) = p_j, where$ $0 \leq j \leq N$
and $\because E(x) = \sum_{k=0}^{N}k*p_k$
$= E(x) = \sum_{k=0}^{N}{\binom{N}{k}a^{N-k}b^{k}k}$
$= E(x) = Nb(a + b)^{N-1}$
(https://www.wolframalpha.com/input/?i=sum+ncr%28n%2C+k%29+*+x%5E%28n-k%29*y%5Ek*k%2Ck%3D0+to+n)
$= E(x) = N\beta(1 – \beta + \beta)$
$= E(x) = N\beta$
$\therefore Answer$ $is$ $B$