In the Black Scholes formula the $N(\alpha)$ gives you cumulative probability, i.e, the probability of a randomly selected occurence being below $\alpha$.
To transform the distribution of your variable into the standard normal you subtract the mean and divide by the standard deviation. It is said in the paragraph preceding formula 13A.5 that the mean is $np$ and the standard deviation is $\sqrt{np(1-p)}$.
So:
- $N(\alpha)$ gives you the cumulative probabily of $\alpha$ in the normal distribution, i.e, probability of a random selection being below $\alpha$
- $N\left(\frac{\alpha - np}{\sqrt{np(1-p)}}\right)$ gives you the cumulative probabily of $\alpha$ in the standard normal distribution
But because what you want is the probability of the random selection being above $\alpha$ (and not below), i.e. $1 - N(x)$, you can use the fact that the normal distribution is symetric and just use $N(-x)$.
Applying this logic to the case above would give you what you want:
$U_2 = N\left(-\frac{\alpha - np}{\sqrt{np(1-p)}}\right) = N\left(\frac{np - \alpha}{\sqrt{np(1-p)}}\right)$
Also, be aware that the price of the binomial model will only converge to the Black Scholes price for a sufficiently larget number of trials.