 Solution

At level $n$, the $N_n$ dyadic cells fully contained in $J$ are disjoint. Their occupied indicators are independent by the preceding part, and every cell has length $2^{-n}$, so each has success probabilityTheir sum therefore satisfiesIf there are no fully contained cells, $N_n=0$ and this means the deterministic zero count. The binomial law follows directly from independence of occupation indicators, without assuming that the full counts $M(D_k^n)$ are already Poisson.