Correct Answer: $90^{\circ}$

Solution :

Given that $ABC$ is an isosceles triangle, $AB = AC$, $AB = AD$ and $\angle ABC = 30^{\circ}$.
Such that,
In $\triangle ABC,$
$\angle ACB =\angle ABC$  (Angles opposite to equal sides of a triangle are also equal)
In $\triangle ACD,$
$\angle ADC =\angle ACD$  (Angles opposite to equal sides of a triangle are also equal)
In $\triangle BCD,$
$⇒\angle BCD=\angle ACB+\angle ACD$
Now, $\angle ABC+\angle BCD+\angle ADC=180^{\circ}$  (Angle sum property)
$⇒\angle ACB+\angle ACB+\angle ACD+ \angle ACD=180^\circ$
$⇒2(\angle ACB+\angle ACD)=180^\circ$
$⇒2\angle BCD=180^{\circ}$
$\therefore\angle BCD=90^{\circ}$
Hence, the correct answer is $90^{\circ}$.