Product of GP

$\begin{array}{rl} a_n &= \ \ 16 \left( \dfrac{1}{4} \right)^{n - 1} \\ \\ \therefore P_n &= \ \ \prod_{k=1}^n 16 \left( \dfrac{1}{4} \right)^{k - 1} \\ \\ &= \ \ 16^n \prod_{k=1}^n \left( \dfrac{1}{4} \right)^{k - 1} \\ \\ &= \ \ 16^n \left( \dfrac{1}{4} \right)^{0 + 1 + 2 + \ldots + (n - 1)} \\ \\ &= \ \ 16^n \left( \dfrac{1}{4} \right)^{\small \displaystyle\sum_{i = 0}^{n - 1} i} \\ \\ &= \ \ 16^n \left( \dfrac{1}{4} \right)^{\Large \frac{n(n -1)}{2}} \\ \\ &= \ \ 2^{4n} \cdot 2^{-n(n - 1)} \\ \\ &= \ \ 2^{n(5 - n)} \\ \\ \implies P_n^{\frac{1}{n}} &= \ \ 2^{5 - n} \\ \\ \therefore \sum_{n = 1}^{\infty} P_n^{\frac{1}{n}} &= \ \ \sum_{n = 1}^{\infty} 2^{5 - n} \\ \\ &= \ \ 2^5 \sum_{n = 1}^{\infty} 2^{-n} \\ \\ &= \ \ 2^5 \left( \dfrac{\frac{1}{2}}{1 - \frac{1}{2}} \right) \\ \\ &= \ \ 2^5 \\ \\ &= \ \ \boxed{32} \end{array}$