One can write the CLT as: $$ \frac{\sum_{i=1}^{n}X_i - n\mu}{\sigma\sqrt{n}} \stackrel{d}{\to} N(0,1) $$
If we consider each $X_i$ here as a Bernoulli random variable that are independent and identically distributed, recalling that the mean and variance of a Bernoulli random variable are $p$ and $p(1-p)$ respectively, we can rewrite this as
$$ \frac{\sum_{i=1}^{n}X_i - np}{\sqrt{np(1-p)}} \stackrel{d}{\to} N(0,1) $$
And then note that a binomial is simply a sum of n Bernoulli random variable. That is that for $X = X_1 + \dots + X_n$ where each $X_i \sim Bernoulli(p)$ we have $X \sim Binomial(n,p)$. So we can finally get the form of the CLT given in your question.
$$ \frac{X - np}{\sqrt{np(1-p)}} \stackrel{d}{\to} N(0,1) $$
This is true even when p is very small, so long as the bernoulli random variables are identically distributed. In the first case where the binomial converges to a poisson distribution, p is growing smaller as n goes to infinity, and so the requirement that the random variables be identically distributed is not satisfied.