Check the three standard facts first.

(A) is true. If $\mathbb{P}(A_1\cap A_2)=\mathbb{P}(A_1)\mathbb{P}(A_2)$ then

$\mathbb{P}(A_1^{c}\cap A_2^{c})=1-\mathbb{P}(A_1\cup A_2) =1-\mathbb{P}(A_1)-\mathbb{P}(A_2)+\mathbb{P}(A_1)\mathbb{P}(A_2) =\bigl(1-\mathbb{P}(A_1)\bigr)\bigl(1-\mathbb{P}(A_2)\bigr).$

(B) is true — this is precisely the second Borel–Cantelli lemma, whose hypothesis of independence is supplied.

(D) is true. If $\mathbb{P}(A)=0$ then $\mathbb{P}(A\cap B)=0=\mathbb{P}(A)\mathbb{P}(B)$ for every $B$ ; if $\mathbb{P}(A)=1$ then $\mathbb{P}(A\cap B)=\mathbb{P}(B)=\mathbb{P}(A)\mathbb{P}(B)$ .

(C) is the odd one out. An event is independent of itself exactly when $\mathbb{P}(A)=\mathbb{P}(A)^{2}$ , i.e. when

$\mathbb{P}(A)\in\{0,1\} .$

So $\mathbb{P}(A)=1$ is sufficient but not necessary — singling out $\mathbb{P}(A)=1$ as the condition is what makes (C) the false statement, since $\mathbb{P}(A)=0$ works equally well. Given that (A), (B) and (D) are all standard true results, (C) is the intended answer.