where $|X|=\frac{3n}{4}$ and $|\bar{X}|=\frac{n}{4}$. Let $$E(G)=\{\{x\}:x\in X\} \cup \{\{x,\bar{x}\}: x\in X \text{ and } \bar{x}\in \bar{X}\}.$$ It is clear that $G_{n}$ contains no closed paths of length $2k$ when $k\geq 1$. Also, $$\begin{aligned}
\lim_{n\to\infty} h_{n}(G_{n}) &= \lim_{n\to\infty} \frac{|X|}{\binom{n}{1}}+\frac{|X|\cdot |\bar{X}|}{\binom{n}{2}} \\
             &= \lim_{n\to\infty} \frac{3}{4}+\frac{\frac{3}{16} n^{2}}{\binom{n}{2}} \\
         &= \frac{3}{4}+\frac{3}{8} =\frac{9}{8}.\end{aligned}$$ Thus $\pi(H)\geq \frac{9}{8}$ for any $H$ containing a closed path of length $2k$ for any $k\geq 1$.

Since $H^{2}$ is bipartite, and $H^{2}$ does not contain a copy of $K_{2}^{\{1,2\}}$, then $H$ is contained in a blow-up of a closed $4$-path. To see this, note that there is a bipartition of the vertices of $H$, $V(H)=A\cup B$, (with respect to the $2$-edges in $H$). Furthermore, we can partition $A$ into $A_{1}\cup A_{2}$ where $v\in A_{1}$ if $\{v\}\in E(H)$ and $v\in A$, $v\in A_{2}$ if $v\in A\setminus A_{1}$. And similarly partition $B$ into $B_{1}\cup B_{2}$ with $v\in B_{1}$ if $\{v\}\in E(H)$ and $v\in B$. Then note that there are no edges from $A_{1}$ to $B_{1}$ since $H$ contains no copy of $K_{2}^{\{1,2\}}$. So $H\subset \bar{P}_{4}(\max\{|A_{1}|, |A_{2}|, |B_{1}|, |B_{2}|)$–a blow-up of $\bar{P}_{4}$. Below is a graphical representation of $H$, illustrating that $H$ is contained in a blow-up of $\bar{P}_{4}$.

(0, 0) circle (1 cm); 