The theorem of total probalility states that
$ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) $
$ P(A|B)=P(A|B \bigcap C)P(C|B)+P(A|B \bigcap C^c)P(C^c|B) $
$ P(A|B)=P(A|B \bigcap C)P(B \bigcap C)/P(B) +P(A|B \bigcap C^c)P(B \bigcap C^c)/P(B) $
Since $ P(A|B)=P(A \bigcap B)/P(B), P(A \bigcap B)=P(A|B)P(B) $
$ P(A|B)P(B)=P(A|B \bigcap C)P(B \bigcap C)+P(A|B \bigcap C^c)P(B \bigcap C^c) $
$ P(A \bigcap B)=P(A \bigcap B \bigcap C)+P(A \bigcap B \bigcap C^C) $
Replacing $ A \bigcap B $ with D
$ P(D)=P(D \bigcap C)+P(D \bigcap C^c) $
$ P(D)=P(D|C)P(C)+P(D|C^c)P(C^c) $
Thus, it is proved.
