Let M be a module over a Noetherian ring A, p a prime ideal of A and (Qi)i∈I a non-empty finite family of submodules of M which are p-primary with respect to M. Then ⋂i∈IQi is p-primary with respect to M.
M/(⋂i∈IQi) is isomorphic to a submodule =0 of the direct sum ⨁i∈I(M/Qi). Now
Ass(i∈I⨁(M/Qi))=i∈I⋃Ass(M/Qi)={p}
(§ 1, no. 1, Corollary 1 to Proposition 3). Hence Ass(M/(⋂i∈IQi))={p} (§ 1, no. 1, Proposition 3 and Corollary 1 to Proposition 2).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.