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Take $c>0$ and $e>0$, as required for the displayed sparse scaling and the later normalization. A strictly balanced graph satisfies $e(F)/v(F)<e/v$ for every proper nonempty subgraph $F\subset H$. This forces $H$ to be connected and to have no isolated vertices: otherwise a component of at least the average density, or deletion of the isolated vertices, violates strict balance. Copies here are subgraphs, not induced subgraphs.

For fixed $r$, the factorial moment ${\mathbb{E}}_r(X_H)$ counts ordered $r$-tuples of distinct copies all present in the binomial random graph. The contribution of vertex-disjoint copies is exactlyThe factor $a^{-r}$ removes the multiple embeddings of each copy coming from its graph automorphisms. For an overlapping tuple, let $U$ be its union. There are $O(n^{v(U)})$ placements of each fixed union type, and the probability its edges are present is $p^{e(U)}$. When adding a copy meeting the earlier union in $f$ vertices and $g$ edges, the exponent $v(U)-(v/e)e(U)$ changes byIt is strictly negative if this intersection is a proper nonempty subgraph, and zero only for an empty intersection or a whole copy. In particular the first vertex-overlapping copy has a proper nonempty intersection: the preceding copies are vertex-disjoint, and a connected copy contained in their union would have to equal one of them, contrary to distinctness. Its contribution is therefore strictly negative; later additions cannot undo that loss. Only finitely many union types occur for fixed $r$, so their total contribution is $o(1)$. We have proved all the factorial moment limits, and the allowed factorial-moment criterion for Poisson convergence givesThis is the Poisson limit for strictly balanced subgraph counts.

The pendant-edge graph has density $7/5$, whereas its proper $K_4$ core has density $6/4$. Thus $K_4^{+}$ is not strictly balanced. The complete graph $K_4$ is strictly balanced, with 24 graph automorphisms, so at $p=c{n}^{-2/3}$ its copy count $Z$ tends to $\mathrm{Po}(c^6/24)$.

Every $K_4^{+}$ copy has a unique four-vertex clique core and one selected edge joining it to a fifth vertex. For any four-set $C$, let $D_C$ count the ambient edges between $C$ and its complement. Then $D_C\sim \mathrm{Bin}(4(n-4),p)$, with mean ${\mu }_n\sim 4c{n}^{1/3}$. It is independent of the internal edges determining whether $C$ is a clique. Choose ${\delta }_n=n^{-1/12}$. A Chernoff bound and a union bound giveTherefore, simultaneously for every actual core, its number of pendant extensions is $(1+o(1))4c{n}^{1/3}$. Extra ambient edges cause no overcount: each selected seven-edge subgraph has exactly one core and one pendant edge. Summing over the cores yieldsbecause $Z$ is uniformly tight. This is the pendant extensions of sparse clique copies mechanism. For $0<\epsilon <1$, neither endpoint of $(k-\epsilon ,k+\epsilon )$ is an integer, and its only nonnegative integer is $k$. Convergence in distribution consequently givesThe positive-$c$ assumption matters: the quotient printed in the question is undefined when $c=0$.