D)
3 out of 4 options can be eliminated with the help of a counter example.
Let $X = \{ a , b , c \}$ and $Y = \{ 1 , 2 \}$
A Function $f$ maps each element of $X$ to exactly one element in $Y$.
Let $f(a)=1 , f(b)=1 , f(c) =1$ and
$A = \{ a \}, B = \{ b , c \}$
A)
$|f ( A \cup B )| = |f (\{a,b,c\})| = |\{1\}| = 1$
$|f(A)|+|f(B)| = 1 + 1 = 2$ , LHS != RHS.
B)
$f ( A \cap B ) = f (\{\}) = \{\}$.
$f(A) \cap f(B) = \{ 1\} \cap \{ 1\} = \{ 1\}$
LHS!=RHS
C)
$|f ( A \cap B )| = |f (\{\})| = |\{ \}| = 0$
$\min\{|f(A)|,|f(B)|\} = \min(1,1) = 1$
LHS!=RHS
D) Its easy to see that this is true because in a function a value can be mapped only to one value. The option assumes inverse of function $f$ exists.