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Use $I_i=1_{A_i}$ and interpret the printed overlap term as $2\Delta ={\sum }_i{\sum }_{j\in J_i}\mathrm{Pr}(A_i\cap A_j)$; its unindexed $A,B$ are a typographical slip. Take $i\notin J_i$, as usual. If necessary, deleting $i$ from its own neighbourhood gives the proof below with a smaller nonnegative right-hand side, so the stated bound still follows.

If $\lambda =0$, every indicator is zero almost surely and the Poisson distribution with parameter zero is the same point mass. Suppose $\lambda >0$. For any $B\subseteq {\mathbb{Z}}_{\ge 0}$, put $h_B(k)=1_{k\in B}$ and let $Z$ have Poisson distribution with mean $\lambda$. Solve the Poisson Stein equationFor example, for $k\ge 1$ one can takeThe full weighted sum is zero, so using its complementary tail also shows this solution is bounded. The allowed Stein factor is $\Vert \Delta g{\Vert }_{\infty }\le \min (1,{\lambda }^{-1})$, where $\Delta g(k)=g(k+1)-g(k)$.

DefineJoint independence of $A_i$ from all events outside its neighbourhood makes $I_i$ independent of $W_i$, so $p_i\mathbb{E}g(W_i+1)=\mathbb{E}{I}_{i}g(W_i+1)$. Adding and subtracting those terms givesA difference over $m$ integer steps has magnitude at most $m\Vert \Delta g{\Vert }_{\infty }$. In the first sum the step count is $I_i+V_i$; in the second, on $I_i=1$, it is $V_i$. Consequently the absolute value is at mostThe Stein equation identifies the left side with $|\mathrm{Pr}(W\in B)-\mathrm{Pr}(Z\in B)|$. Taking the supremum over $B$ is the definition of total variation distance, proving the Stein-Chen bound with the Poisson Stein factor:

For the random uniform hypergraph, index indicators by four-vertex sets. Each such set is a hypergraph clique exactly when its four triples are all present, so its probability is $p^4$. Distinct sets use disjoint triple indicators unless their intersection has size three; disjoint underlying trials give the required joint independence. A fixed four-set has $d=4(n-4)$ dependent others, and each overlapping pair requires seven distinct triples, with joint probability $p^7$. Thus, with $N=\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ and $\lambda =N{p}^4$, the bracket isSince $\lambda \min (1,{\lambda }^{-1})\le 1$,This calculation applies for $n\ge 4$ and $p>0$; if $p=0$ or $n<4$, the count and its Poisson parameter are zero and the conclusion is immediate.