Everything follows from the first Borel–Cantelli lemma, which needs no independence:

$\sum_{n\ge1}\mathbb{P}(A_n)<\infty \;\Longrightarrow\; \mathbb{P}(\limsup A_n)=\mathbb{P}[A_n \text{ i.o.}]=0 .$

(D) is true, being exactly the contrapositive of (BC1): if the series converged, $\mathbb{P}(\limsup A_n)$ would be $0$ , not $1$ .

(B) is false. Without independence the divergence of $\sum\mathbb{P}(A_n)$ says nothing. Take $\Omega=(0,1)$ with Lebesgue measure and $A_n=(0,\tfrac1n)$ : then $\sum\mathbb{P}(A_n)=\sum\tfrac1n=\infty$ , yet $\liminf A_n=\limsup A_n=\varnothing$ , so the probability is $0$ , not $1$ . (Even the second Borel–Cantelli lemma, which does assume independence, concludes about $\limsup$ , not $\liminf$ .)

(C) is false — it is the converse of (BC1). The same sequence $A_n=(0,\tfrac1n)$ has $\mathbb{P}(\limsup A_n)=0$ while $\sum\mathbb{P}(A_n)=\infty$ .

Note — the question admits two correct options. Statement (A) is also always true: (BC1) gives $\mathbb{P}(\limsup A_n)=0$ , and since $\liminf A_n\subseteq\limsup A_n$ we get $\mathbb{P}(\liminf A_n)=0$ as well. So (A) and (D) are both valid, and the item is defective. (D) is the cleaner statement of the standard result and is taken here as the key; a candidate who marked (A) has an equally defensible answer.