It is given in the question that:
$1.$ $AB → CD$
$2.$ $AF → D$
$3.$ $DE → F$
$4.$ $C → G$
$5.$ $F → E$
$6.$ $G → A$
We will go through each option one by one.
Option A: $\left \{C F\right \}^{*}$ = $\left \{ACDEFG\right \}$
$=>$ $\left \{C F\right \}^{*}$ = $\underbrace{\left \{CGA\right \}}_\textrm{Closure of C using 4 and 6} + \underbrace{\left \{FDE\right \}}_\textrm{Closure of F using 2 and 5}$
$=>$ $\left \{C F\right \}^{*}$ = $\left \{CGAFDE\right \}$
If we just rearrange the terms:
$=>$ $\left \{C F\right \}^{*}$ = $\left \{ACDEFG\right \}$
Hence $\color{green} True$
Option B: $\left \{BG\right \}^{*}$ = $\left \{ABCDG\right \}$
$=>$ $\left \{BG\right \}^{*}$ = $\underbrace{\left \{GA\right \}}_\textrm{Closure of G using 6} + \underbrace{\left \{BCD\right \}}_\textrm{Closure of B using 1: Note that A from G’s closure is being used}$
$=>$ $\left \{BG\right \}^{*}$ = $\left \{GABCD\right \}$
If we just rearrange the terms:
$=>$ $\left \{BG\right \}^{*}$ = $\left \{ABCDG\right \}$
Hence $\color{green} True$
Option C: $\left \{AF\right \}^{*}$ = $\left \{ACDEFG\right \}$
$=>$ $\left \{AF\right \}^{*}$ = $\underbrace{\left \{A\right \}}_\textrm{Closure of A} + \underbrace{\left \{FDE\right \}}_\textrm{Closure of F using 2, 3 and 5: Note that A from A’s closure is being used}$
$=>$ $\left \{AF\right \}^{*}$ = $\left \{AFDE\right \}$
If we just rearrange the terms:
$=>$ $\left \{AF\right \}^{*}$ = $\left \{ADEF\right \}$
But this is not equal to $\left \{AF\right \}^{*}$ = $\left \{ACDEFG\right \}$
Hence $\color{red} False$
Option D: $\left \{AB\right \}^{*}$ = $\left \{ABCDG\right \}$
$=>$ $\left \{AB\right \}^{*}$ = $\underbrace{\left \{ABCD\right \}}_\textrm{Closure of AB using 1} + \underbrace{\left \{G\right \}}_\textrm{Closure of C using 4: Note that C from AB’s closure is being used}$
$=>$ $\left \{AB\right \}^{*}$ = $\left \{ABCDG\right \}$
Hence $\color{green} True$
Hence only False option C (Answer).