= Solution
Let $\sigma_N=\sqrt{Np(1-p)}$ and use the <floor function> to choose
$$
\boxed{k_a(N)=\max\{0,\lfloor Np+a\sigma_N\rfloor+1\},\qquad
k_b(N)=\min\{N,\lfloor Np+b\sigma_N\rfloor\}.}
$$
Interpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/\sigma_N\leq b$, where $K_N$ has the <binomial distribution> of parameters $N,p$. Since $0<p<1$, the <central limit theorem> for a sum of <independent> <Bernoulli distribution> variables gives <convergence in distribution> of that standardized count to the <standard normal distribution>. Its <cumulative distribution function> is continuous at both endpoints, so
$$
\boxed{\sum_{k=k_a(N)}^{k_b(N)}p_k(N,p)
\longrightarrow\Phi(b)-\Phi(a)
=\frac1{\sqrt{2\pi}}\int_a^be^{-u^2/2}\,du.}
$$
The TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.
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