The mean $\mu$ of a binomial = np. The standard deviation of a binomial = $\sqrt{np(1-p)}$
For a normal distribution, $\mu$ should be 3 standard deviations away from 0 and n.
Therefore:
$\mu$ - $3\sqrt{np(1-p)} > 0 \hspace{2cm}$ and $\hspace{2cm}\mu$ + $3\sqrt{np(1-p)}<n$
From that starting point, algebraically you can get to the inequalities:
$np>9(1-p)\hspace{2cm}$ and $\hspace{2cm}n(1-p)>9p$
To satisfy these inequalities, as n gets larger, p has a wider range. Or you could also say the closer p is to 0.5, the smaller n you can use.
Using n=10 (for example):
$0.474<p<0.526$
As n gets larger, p does not have to be so close to 0.5. For n = 100,
$0.0826<p<0.9174$
Remarkably, even with a p = 0.9, if n >100 then the mean will be 3 standard deviations away from 0 and n.
This relates to calculating np and n(1-p), as if both are greater than 5, usually these inequalities are satisfied. However something like n=15, p=0.65 does not work, so some textbooks say np>9.
This condition does not guarantee that the binomial will fit a normal dist. but just that the mean will not be skewed too far towards 0 or n.
Best Answer
From what i've been taught, the aproximation is valid when: \begin{equation*} X \sim B(n,p) \Rightarrow X \sim N(np,np(1-p)) \text{ (aproximated) iff}\hspace{.2cm} n\geq 20, np > 5 \text{ and } n(1-p)>5 \end{equation*} Hope it helped.