= Solution
For each $i$, the <total variation distance> between $\operatorname{Bernoulli}(p_i)$ and $\operatorname{Poisson}(p_i)$ is
$$
p_i(1-e^{-p_i})\leq p_i^2.
$$
Indeed, compare their masses at zero, one, and the Poisson tail; the positive excess of Bernoulli mass at one is $p_i(1-e^{-p_i})$. A maximal coupling therefore gives a pair $(X_i,N_i)$ with mismatch probability at most $p_i^2$. Couple these pairs independently. Then
$$
\mathbb P\left(\sum_iX_i\ne\sum_iN_i\right)
\leq\sum_i p_i^2
$$
by the union bound. Since $\sum_iN_i\sim\operatorname{Poisson}(\sum_ip_i)=\operatorname{Poisson}(\nu)$, part (b) yields
$$
d_{\mathrm{TV}}(P,Q)\leq\sum_{i=1}^np_i^2.
$$
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