I have done the following :
We have that $(A\cap B)\cap (A^c\cap B)=\emptyset$ and so $$P((A\cap B)\cup (A^c\cap B))=P(A\cap B)+P (A^c\cap B) \\ \Rightarrow P((A\cup A^c)\cap B)=P(A\cap B)+P (A^c\cap B)$$
Now we have to show that $P((A\cup A^c)\cap B)=P(A)P(B)+P(A^c)P(B)=[P(A)+P(A^c)]P(B)$, right?
We get that result if $(A\cup A^c)$ and $B$ are independent, or not? How can we show that?
Or is there an other (better) way to show the desired expression?