Unfortunately, there seems to be a counterexample: Let $B^p_i$ be the interval $(p-i,\infty)$ in the integers $\mathbb{Z}$, for natural numbers $p$ and $i$. In this case, we have $B^{p+1}_i\subset B^p_i$, and for any fixed $i$, we have $\bigcap_p B^p_i=0$, since $p$ runs out to infinty. But for fixed $p$, the limit of $B^p_i$ as $i$ increases is all of $\mathbb{Z}$. Thus, the intersection of these limits is also $\mathbb{Z}$, which is not empty.
A simpler counterexample, using the same idea: let $B^p_i=\{1\}$B^p_i=A$ for some fixed non-empty $A$, if $p\lt i$, otherwise $B^p_i=$ emptyset. In this case, for fixed $i$ we have $\bigcap_p B^p_i=0$, since eventually $p$ exceeds $i$, but for fixed $p$, the $B^p_i$ are eventually equal to {1} $A$ as $i$ increases, and so have limit {1}.$A$.
