(0,-3) circle (1 cm); (3, 0) circle (1 cm); (3,-3) circle (1 cm); at (-1.5, 0) [$A_{1}$]{}; at (-1.5,-3) [$A_{2}$]{}; at (4.5, 0) [$B_{2}$]{}; at (4.5, -3) [$B_{1}$]{};

at (0, .5) \[vertex\_closed\] (v1) ; at (-.5, -.5) \[vertex\_closed\] (v2) ; at (.35,-.35) \[vertex\_closed\] (v3) ;

at (3, .5) \[vertex\_open\] (v4) ; at (2.65,-.35) \[vertex\_open\] (v5) ; at (3.5, -.5) \[vertex\_open\] (v6) ;

at (0, -3.5) \[vertex\_open\] (v7) ; at (-.5, -2.65) \[vertex\_open\] (v8) ;

at (3, -3.5) \[vertex\_closed\] (v9) ; at (3.5, -2.65) \[vertex\_closed\] (v10) ;

(.6,-.2)–(2.75, 0); (.35, .2)–(2.5, .5); (3,-.4)–(0,-2.65); (3.15, -.5)–(.35, -3); (.5, -3.5)–(2.5, -3.25); (.65,-3.25)–(2.7, -2.8);

Since $\pi(H)\leq \pi(\bar{P}_{4}(s))=\pi(\bar{P}_{4})$ we need only show that $\pi(\bar{P}_{4})\leq \frac{9}{8}$. Let $G_{n}$ be a family of $\bar{P}_{4}$-free graphs such that $h_{n}(G_{n})=\pi_{n}(\bar{P}_{4})$. Partition the vertices of $G_{n}$ as follows: $$\begin{aligned}
X_{n} &= \{v:\{v\}\in E(G_{n})\}, \\
Y_{n} &= \{v: \{v\}\notin E(G_{n}) \text{ and } \exists x_{1}\neq x_{2}\in X_{n} \text{ with } \{x_{1}, v\}, \{x_{2}, v\} \in E(G_{n})\}, \\
Z_{n} &= V(G)\setminus (X_{n}\cup Y_{n}).\end{aligned}$$ Let us say that $|X_{n}|=xn$, $|Y_{n}|=yn$ and hence $|Z_{n}|=(1-x-y)n$.

First, note that $E(G)\cap \binom{Y_{n}}{2}=\emptyset$. Otherwise, since each vertex in $Y_{n}$ has at least 2 neighbors in $X_{n}$, $G_{n}$ would contain a closed path of length $4$. Also, each vertex 