= Solution
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 exactly
$$
\frac{(n)_{rv}}{a^r}p^{re}\longrightarrow\left(\frac{c^e}{a}\right)^r.
$$
The 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 by
$$
-f+(v/e)g\le0.
$$
It 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> gives
$$
\boxed{X_H(G(n,cn^{-v/e}))\Rightarrow\operatorname{Po}(c^e/a).}
$$
This 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 \b[$K_4^+$ is not strictly balanced]. The <complete graph> $K_4$ is strictly balanced, with 24 <graph automorphisms>, so at $p=cn^{-2/3}$ its copy count $Z$ tends to $\operatorname{Po}(c^6/24)$.
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 $D_C\sim\operatorname{Bin}(4(n-4),p)$, with <mean> $\mu_n\sim4cn^{1/3}$. 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> give
$$
\Pr\bigl(\exists C:|D_C-\mu_n|>\delta_n\mu_n\bigr)\le2\binom n4\exp(-\delta_n^2\mu_n/3)=o(1).
$$
Therefore, simultaneously for every actual core, its number of pendant extensions is $(1+o(1))4cn^{1/3}$. Extra ambient <edges> cause no overcount: each selected seven-edge <subgraph> has exactly one core and one pendant <edge>. Summing over the cores yields
$$
\frac{X_{K_4^+}}{4cn^{1/3}}=(1+o_{\mathbb P}(1))Z,\qquad \frac{X_{K_4^+}}{4cn^{1/3}}-Z\xrightarrow{\mathbb P}0,
$$
because $Z$ is <uniformly tight>. This is the <pendant extensions of sparse clique copies> mechanism. For $0<\varepsilon<1$, neither endpoint of $(k-\varepsilon,k+\varepsilon)$ is an integer, and its only nonnegative integer is $k$. <Convergence in distribution> consequently gives
$$
\boxed{\Pr\left(k-\varepsilon<\frac{X_{K_4^+}}{4cn^{1/3}}<k+\varepsilon\right)\longrightarrow e^{-c^6/24}\frac{(c^6/24)^k}{k!}.}
$$
The positive-$c$ assumption matters: the quotient printed in the question is undefined when $c=0$.
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