We know that a binomial distribution can be approximated with a normal distribution when $n * p$ is large where $n$ and $p$ are the number of bernoulli trials and $p$ is the probability of success of a bernoulli trial. So binom(n, p) can be approximated with $N(np, np(1-p))$
I am wondering if we can somehow do the reverse? i.e., given a normal distribution $N(\mu, \sigma^2)$, can we find parameters $n$ and $p$ and approximate the normal distribution with a binomial distribution?
You can always just solve for $n$ and $p$ in terms of $\mu$ and $\sigma^2$. Given a normal distribution $N(\mu, \sigma^2)$, set $p = 1 - \frac{\sigma^2}{\mu}$ and $n = \frac{\mu}{1-\sigma^2/\mu}$ (if this isn't an integer, just round it). Then you know that the binomial distribution $\operatorname{Binom}(n,p)$ will be approximately equal to $N(np, np(1-p)) \approx N(\mu, \sigma^2)$.
EDIT: Also this is only possible if $\sigma^2 < \mu$, else you would get $p<0$.