Refer Attached Image

In triangle ABC, AB = AC = 5, that means its an isosceles triangle. This question is based on a property of isosceles triangle.
"In an Isosceles Triangle the perpendicular from the vertex to the non equal side is a perpendicular bisector to the non-equal side"

In the attached figure, AD which is a perpendicular draw from A to BC will touch BC at its mid-point, which is D. => BD = DC = 8/2 = 4
To find the Area of ABC we need to find the height of ABC which is AD.

Using Pythagoras theorem on ADB we get
$AB^2$ =$BD^2$ + $AD^2$
=> $5^2$ =$4^2$ + $AD^2$
=> $AD^2$ = 25 - 16 = 9 = $3^2$
=> AD = 3

So, Area of ABC = $\frac{1}{2}$BC * AD = $\frac{1}{2}$ * 8 * 3 = 12

So, Answer will be B
Hope it helps!