Apply the distributive laws as Graham Kemp enunciated to get
$$(A \wedge B \wedge M) \vee (\neg F \wedge B)
= ((A \wedge B \wedge M) \vee \neg F) \wedge ((A \wedge B \wedge M) \vee B)$$
Now from distributivity again,
$$(A \wedge B \wedge M) \vee \neg F
= (A \vee \neg F) \wedge (B \vee \neg F) \wedge (M \vee \neg F),$$
while absorption, commutativity and associativity yield
$$(A \wedge B \wedge M) \vee B = B,$$
and so we get the expression
$$(A \vee \neg F) \wedge (B \vee \neg F) \wedge (M \vee \neg F) \wedge B.$$
Again, by absorption,
$$(B \vee \neg F) \wedge B = B,$$
and so the final expression, in CNF is
$$(A \vee \neg F) \wedge (M \vee \neg F) \wedge B.$$