Answer
$P(A)=\frac{1}{4}$.
$P(C|B=empty)=P(D|B=empty)=\frac{3}{8}.$
Work Step by Step
Suppose the doors are $A,B,C$ and $D$. Without Loss of Generality, assume you choose door $A$, and the game-show host opened the door $B$ to reveal it does not contain the prize. Now if you do not change then your chance of winning is same as before opening door $B$ i.e., $P(A)=\frac{1}{4}$.
However if you change, then $P(C \cup D|B=empty)=P(A^c)=\frac{3}{4}$.
Since the prize being either of $C$ or $D$ is equally likely, hence $P(C|B=empty)=P(D|B=empty)=\frac{3}{8}.$
