Can someone explain how the final equation is obtained? I cannot understand how the exponential term occurs. from Mackay Pg. 17
$1=\sum_{K}\left(\begin{array}{l}N \\ K\end{array}\right) 2^{-N} \simeq 2^{-N}\left(\begin{array}{c}N \\ N / 2\end{array}\right) \sum_{r=-N / 2}^{N / 2} e^{-r^{2} / 2 \sigma^{2}} \simeq 2^{-N}\left(\begin{array}{c}N \\ N / 2\end{array}\right) \sqrt{2 \pi} \sigma$
where $\sigma$ = $\sqrt{N/4}$ from $\left(\begin{array}{c}N \\ N / 2\end{array}\right) \simeq 2^{N} \frac{1}{\sqrt{2 \pi N / 4}}$
It's a rather awkward derivation of the asymptotic value of the central binomial coefficient.
I would put in a slightly simpler form: We know that $p_k = \binom{N}{k} 2^{-N}$ corresponds to a Binomial $(N,\frac12)$ probability distribution, which is the sum of $N$ fair Bernoulli variables. For large $N$, the CLT applies, and $p_k$ can be approximated by a gaussian $g_k =\frac{1}{\sqrt{2 \pi \sigma^2}} \exp( -\frac12(k- \mu)^2)$ with mean $\mu=N/2$ and variance $ \sigma^2=N/4$.
Now, we evaluate that approximation for $k=N/2=\mu$:
$$ \binom{N}{N/2} 2^{-N} \approx \frac{1}{\sqrt{2 \pi \sigma^2}} = \frac{1}{\sqrt{2 \pi N / 4}} $$