$\newcommand{\ep}{\varepsilon}
\newcommand{\Si}{\Sigma}$The answer is yes, and it is enough that $N_k\to\infty$ just in probability or, equivalently, in distribution (rather than almost surely), which means that for each real $b$
$$P(N_k\le b)\to0$$
as $k\to\infty$.
Indeed, let $F_n$ be the cdf of $\sum_1^n(X_i-\mu)/(\sigma\sqrt n)$ and let $G_k$ be the cdf of $S_k$.
Take any real $x$. By the central limit theorem, $F_n(x)\to\Phi(x)$ as $n\to\infty$, where $\Phi$ is the standard normal cdf.
Take now any real $\ep>0$. Then there is some natural $A$ such that $|F_n(x)-\Phi(x)|<\ep/2$ for all $n>A$. Further, there is some natural $K$ such that $P(N_k\le A)<\ep/2$ for all $k>K$.
Take now any such $A$ and $K$, and then take any natural $k>K$. Write
$$G_k(x)-\Phi(x)=\sum_{n=1}^\infty P(N_k=n)F_n(x)-\Phi(x)=\Si_1+\Si_2,
$$
where
$$|\Si_1|=\Big|\sum_{n=1}^A P(N_k=n)(F_n(x)-\Phi(x))\Big|
\le\sum_{n=1}^A P(N_k=n)=P(N_k\le A)<\ep/2,$$
$$|\Si_2|=\Big|\sum_{n>A} P(N_k=n)(F_n(x)-\Phi(x))\Big|
\le\sum_{n>A} P(N_k=n)\ep/2\le\ep/2,$$
whence
$$|G_k(x)-\Phi(x)|<\ep$$
for all $k>K$. That is, $G_k(x)\to\Phi(x)$ as $k\to\infty$, as desired.