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By symmetry I can assume $p\le\frac12$. I will use natural logs.

Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$.

The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$.

In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\ge (2+\epsilon)\ln\ln n/n$.

In the range $k_1+1 \le k\le n$, $b_k\le 2^{-3n/4}$ (using the assumption $p\le \frac12$) and $f(x)-f_j(x)=O(p^{-3})$, so this part is negligible compared to the previous part.

In summary, $$\sum_{k=0}^n b_k f(k/n) = \frac{1+o(1)}{2n}$$ for $(2+\epsilon)\ln\ln n/n\le p\le 1-(2+\epsilon)\ln\ln n/n$, any $\epsilon\ge 0$.

For the case $p=a/n$ with $a=O(\ln\ln n)$, use $\binom{n}{k}=\frac{n^k}{k!}\left(1-\binom{k}{2}/n -O(k^3/n^2)\right)$ and simple bounds on the tail to find $$\sum_{k=0}^n b_k f(k/n) = \sum_{k=0}^{(\ln n)^2} \frac{a^k}{k!}\left(1-\frac{\binom{k}{2}}{n}-\frac{ak}{n}\right)f(k/n) + O((\ln n)^{O(1)}/n^2).$$

By symmetry I can assume $p\le\frac12$. I will use natural logs.

Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$.

The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$.

In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\ge (2+\epsilon)\ln\ln n/n$.

In the range $k_1+1 \le k\le n$, $b_k\le 2^{-3n/4}$ (using the assumption $p\le \frac12$) and $f(x)-f_j(x)=O(p^{-3})$, so this part is negligible compared to the previous part.

In summary, $$\sum_{k=0}^n b_k f(k/n) = \frac{1+o(1)}{2n}$$ for $(2+\epsilon)\ln\ln n/n\le p\le 1-(2+\epsilon)\ln\ln n/n$, any $\epsilon\gt 0$.

For the case $p=a/n$ with $a=O(\ln\ln n)$, use $\binom{n}{k}=\frac{n^k}{k!}\left(1-\binom{k}{2}/n -O(k^3/n^2)\right)$ and simple bounds on the tail to find $$\sum_{k=0}^n b_k f(k/n) = \sum_{k=0}^{(\ln n)^2} \frac{a^k}{k!}\left(1-\frac{\binom{k}{2}}{n}-\frac{ak}{n}\right)f(k/n) + O((\ln n)^{O(1)}/n^2).$$

By symmetry I can assume $p\le\frac12$. I will use natural logs.

Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$.

The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$.

In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\ge (2+\epsilon)\ln\ln n/n$.

In the range $k_1+1 \le k\le n$, $b_k\le 2^{-3n/4}$ (using the assumption $p\le \frac12$) and $f(x)-f_j(x)=O(p^{-3})$, so this part is negligible compared to the previous part.

In summary, $$\sum_{k=0}^n b_k f(k/n) = \frac{1+o(1)}{2n}$$ for $(2+\epsilon)\ln\ln n/n\le p\le 1-(2+\epsilon)\ln\ln n/n$, any $\epsilon\ge 0$.

By symmetry I can assume $p\le\frac12$. I will use natural logs.

Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$.

The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$.

In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\ge (2+\epsilon)\ln\ln n/n$.

In the range $k_1+1 \le k\le n$, $b_k\le 2^{-3n/4}$ (using the assumption $p\le \frac12$) and $f(x)-f_j(x)=O(p^{-3})$, so this part is negligible compared to the previous part.

In summary, $$\sum_{k=0}^n b_k f(k/n) = \frac{1+o(1)}{2n}$$ for $(2+\epsilon)\ln\ln n/n\le p\le 1-(2+\epsilon)\ln\ln n/n$, any $\epsilon\ge 0$.

For the case $p=a/n$ with $a=O(\ln\ln n)$, use $\binom{n}{k}=\frac{n^k}{k!}\left(1-\binom{k}{2}/n -O(k^3/n^2)\right)$ and simple bounds on the tail to find $$\sum_{k=0}^n b_k f(k/n) = \sum_{k=0}^{(\ln n)^2} \frac{a^k}{k!}\left(1-\frac{\binom{k}{2}}{n}-\frac{ak}{n}\right)f(k/n) + O((\ln n)^{O(1)}/n^2).$$

By symmetry I can assume $p\le\frac12$. I will use natural logs.

Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$.

The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$.

In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\gt (2+\epsilon)\ln\ln n/n$.

TO BE CONTINUED

By symmetry I can assume $p\le\frac12$. I will use natural logs.

Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$.

The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$.

In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\ge (2+\epsilon)\ln\ln n/n$.

In the range $k_1+1 \le k\le n$, $b_k\le 2^{-3n/4}$ (using the assumption $p\le \frac12$) and $f(x)-f_j(x)=O(p^{-3})$, so this part is negligible compared to the previous part.

In summary, $$\sum_{k=0}^n b_k f(k/n) = \frac{1+o(1)}{2n}$$ for $(2+\epsilon)\ln\ln n/n\le p\le 1-(2+\epsilon)\ln\ln n/n$, any $\epsilon\ge 0$.