Distribution of $\sqrt{npq}(X-np)$ for large n where $X$ has a binomial distribution with parameters $n,p$ and $q=1-p$
Solution 1:
In fact $a=\dfrac{1}{\sqrt{npq}}$ because the "normalizing" formula is
$$Y=\dfrac{X-m}{\sigma}$$
with mean $m=np$ and standard deviation $\sigma=\sqrt{npq}$ for a Bin(n,p) distribution.