Let 'a' be the side of the square
ABCD is a square of $a^{2}$ = 4 $\Rightarrow$ a = 2.
AC = BD = $2\sqrt{2}$
perimeters of four triangles = AB + BC + CD + DA + 2(AC + BD) = 8 + 2 ($2\sqrt{2}$ + $2\sqrt{2}$)
= 8 ( 1 + $2\sqrt{2}$)
Option A is the Correct Answer.