Study the sequence $x_n=\sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}$.

Let $$ a_n=\max\{\sin k: k=1,\ldots,n\}. $$ Then $\{a_n\}$ is increasing and $a_n\le 1$, for all $n\in\mathbb N$, and hence convergent to some positive $a\le 1$. (In fact, it converges to 1.)

Then $$ 2^{na_n}\le 2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}\le n\cdot 2^{na_n} $$ and hence $$ 2^{a_n}= \sqrt[n]{2^{na_n}}\le \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}\le 2^{a_n}\sqrt[n]{n} $$ But $\,\,2^{a_n}\to 2^a$ and $\,\,\sqrt[n]{n}2^{a_n}\to 2^a$, and hence $$ \sqrt[n]{2^{na_n}}\le \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}} \to 2^a. $$ Thus A is the correct answer.

Note. In fact $a=1$, and thus $$ \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}\to 2. $$ This is due to Weyl's Equidistribution Criterion.

not a solution

Here is $n=2$ to $n=1000$.

Visual impression: converges to $2$, but not monotonically.