Let $S$ be the set of bitstrings of length $n$ with $pn$ ones and $(1-p)n$ zeros, for some $p \in [0,1]$. I want to show that, for large $n$, if one uniformly samples pairs from $(s,s') \in S^2$, then $$ \mathbb{E}(d(s,s')) \approx \frac{n}{2}H(p), \tag{1} $$ where $d$ is the Hamming distance and $H(p) = -p\log_2(p) -(1-p)\log_2(1-p)$ is the binary entropy. Since $\binom{n}{pn} \approx 2^{nH(p)}$, showing that $\mathbb{E}(d(s,s')) \approx \log(\binom{n}{pn})$ would also do the job.
My current approach has brought me this far: We have that $|S| \approx 2^{nH(p)}$. Let $T \equiv \{0,1\}^{nH(p)}$. If we uniformly sample from $(t,t')\in T^2$, then $$ \mathbb{E}(d(t,t')) = \frac{n}{2}H(p). $$ This is because this latter process can be described as a Bernoulli trial and the Hamming distance (as a random variable) follows a binomial distribution $B(nH(p),1/2)$. Now, clearly any bijective map $B: T \to S$, induces a uniform distribution over $S^2 = B(T)^2$ by sampling uniformly from $T^2$. Hence, for any $B$, $$ \mathbb{E}(d(B(t),B(t'))) = \mathbb{E}(d(s,s')). $$ This is because they all induce the same process (correct?). Consequently, we only need to find a single $B$ such that $$ d(B(t),B(t')) \approx d(t,t'), \quad \forall t,t' \in T \tag{2} $$ to conclude (1). Does a bijection that satisfies (2) exist (maybe it can be constructed on the basis of Gray codes)? Or maybe a completely different, simpler way to show (1)?
The simplest way to compute the expected Hamming distance is to use linearity of expectation, which shows that $$ \operatorname*{\mathbb{E}}_{x,y \in S}[d(x,y)] = n\Pr_{x,y \in S}[x_1 \neq y_1] = 2np(1-p). $$ As a sanity check, if $p=0$ or $p=1$ then the expected distance is 0, whereas if $p=1/2$ then the expected distance is $n/2$.