Given that, $4^{\text{th}}$ term of an arithmetic progression is $8.$
$\Rightarrow \boxed{T_{4} = 8}$
We know that, $n^{\text{th}}$ term of an arithmetic progression $: T_{n} = a + (n-1)d$
Where $a =$ the first term, $d =$ common difference.
Now, $T_{4} = a + (4-1)d$
$ \Rightarrow 8 = a + 3d$
$ \Rightarrow \boxed{a + 3d = 8}$
The sum of an arithmetic progression $: S_{n} = \frac{n}{2} \left[ 2a + (n-1) d \right]$
$\Rightarrow S_{7} = \frac{7}{2} \left[ 2a + (7-1) d \right]$
$\Rightarrow S_{7} = \frac{7}{2} (2a + 6d)$
$\Rightarrow S_{7} = \frac{7}{2} \times 2 (a + 3d)$
$\Rightarrow S_{7} = 7 \times 8 \quad [\because a + 3d = 8]$
$\Rightarrow \boxed{S_{7} = 56}$
$\therefore$ The sum of the first $7$ terms of the arithmetic progression is $56.$
Correct Answer $: \text{C}$