A construction that uses the equality $\kappa^\lambda=\kappa$ directly runs as follows. By that equality the set $H$ has cardinality $\kappa$, so we can find a surjection $f:\kappa\to H$ such that for every $\varepsilon\in H$ the set $\{\alpha:f(\alpha)=\varepsilon\}$ has cardinality $\kappa$, and $\operatorname{dom}\varepsilon\subseteq \alpha$ whenever $f(\alpha)=\varepsilon$. Note that the cofinality of $\kappa=2^\lambda$ is larger than $\lambda$, hence the domains of the members of $H$ are bounded in $\kappa$.

Now let $G=\bigl\{\{\beta,\alpha\}:\beta\in\operatorname{dom}f(\alpha)$ and $f(\alpha)(\beta)=0\bigr\}$. Then for every $\varepsilon\in H$ the set $G_\varepsilon$ contains $\{\alpha:f(\alpha)=\varepsilon\}$, so it has cardinality $\kappa$.