$\displaystyle\lim_{k\rightarrow \infty}\left(\sum_{i=1}^{n}\frac{1}{|x_{i}|^{k}}\right)^{-1/k} $
$ =\displaystyle\lim_{k \to \infty} \left(\sum_{i} n_i\frac{1}{|a_i|^k}\right)^{-1/k}$ (The $i$ here is different from the previous expression and denote the $i$ unique real values repeated $n_i$ times each.
$=\displaystyle\lim_{k \to \infty} \left(\sum_{i} n_i |a_i|^{-k}\right)^{-1/k}$
Since, $k$ is very large $(\to \infty),$ the summation above will be dominated by the largest value for $n_i |a_i|^{-k}$ which will come for $\min |a_i |$ due to $-k$ in exponent. We can also say that the summation series converges to $n_i {(\min |a_i|)}^{-k}.$ So, our limit becomes
$=\displaystyle \lim_{k \to \infty} \left({n_i {(\min |a_i|)}^{-k}}\right)^{-1/k} = n_i^{-1/k} ((\min |a_i|)^{-k})^{-1/k}$
$\quad = n_i^0 \min |a_i| = \min |a_i|$
Correct Option: B.