Consider $\triangle ABC$
We are given that $AB = BC = CD$

We can infer from the diagram that $AB$ and $AC$ are both radii of the circles
So, now we have $AB = AC = r$ also $AB = BC$

Hence, we can easily say that $AB = BC = AC$, which makes $\triangle ABC$ an equilateral triangle

Using property - All angles in a equilateral triangle are equal to 60 degrees

Similar case goes for $\triangle ADC$ which essentially makes $BD$ an median and angle bisector at $B$ as well as at $D$

Therefore, $x^o = \frac{∠ABC}{2} = \frac{60^o}{2} = 30^o$

Hence, Answer is B