Answer
$P(A\cup B\cup C\cup D)=P(A)+P(B)+P(C)+P(D)-3P(A\cap B\cap C \cap D)$
Work Step by Step
The intersection between the four events is such that:
$P(A\cap B\cap C \cap D)=P(A\cap B)=P(A\cap C)=P(A\cap D)=P(B\cap C)=P(B\cap D)=P(C\cap D)=P(A\cap B\cap C)=P(A\cap B\cap D)=P(A\cap C\cap D)=P(B\cap C\cap D)$
So, when we add $P(A)+P(B)+P(C)+P(D)$ we add the intersection $P(A\cap B\cap C \cap D)$ four times. So, it needs to be removed three times.