Answer: Option B
According to question,
Let ∠CAB = x and ∠CBA = y
$$\eqalign{
& \Rightarrow \angle CAD = \frac{{180 - x}}{2} = 90 - \frac{x}{2} \cr
& {\text{and}} \cr
& \Rightarrow \angle EBC = \frac{{180 - y}}{2} = 90 - \frac{y}{2} \cr
& {\text{also}}\,\angle AEB = \angle EAB = x \cr} $$
(∵ AB = EB ⇒ ABE is an isosceles triangle)
and ∠ADB = ∠ABD = y
(∵ AB = AD ⇒ ADB is an isosceles triangle)
In ΔAEB,
∠AEB + ∠ABE + ∠BAE = 180°
x + x + y + 90 - $$\frac{y}{2}$$ = 180°
⇒ 4x + y = 180°
Similarly in ΔADB
4y + x = 180°
⇒ 4y + x + 4x + y = 180 + 180
⇒ 5x + 5y = 360°
⇒ x + y = 72°
In triangle ABC,
∠ACB + x + y = 180°
⇒ ∠ACB = 180 - 72
⇒ ∠ACB = 108°