This problem involves the quadrilateral $ABCD$, where $AB = BC = AD$ and $\angle DAB = 90^\circ$. The given angles in the figure are $\angle ABC = 60^\circ$ and $\angle CDA = x$. We are asked to find the value of $x$.

Step-by-step Solution:

1. Identify the shape: Since $AB = BC = AD$ and $\angle DAB = 90^\circ$, the quadrilateral $ABCD$ forms an isosceles triangle for $\triangle ABC$ and a quadrilateral with specific angle conditions.

2. Sum of angles in a triangle: In $\triangle ABC$, the sum of the angles must equal $180^\circ$. $\angle ABC + \angle ACB + \angle BAC = 180^\circ$ We know $\angle ABC = 60^\circ$ and $\angle BAC = 90^\circ$, so: $60^\circ + \angle ACB + 90^\circ = 180^\circ$ Simplifying this: $\angle ACB = 30^\circ$

3. Using angle sum in quadrilateral: The sum of all internal angles of any quadrilateral is $360^\circ$. $\angle DAB + \angle ABC + \angle BCD + \angle CDA = 360^\circ$ Substituting the known values: $90^\circ + 60^\circ + 30^\circ + x = 360^\circ$ Simplifying: $180^\circ + x = 360^\circ$ $x = 180^\circ$

4. Final value of $x$: The value of $x = 180^\circ$, but this seems incorrect based on the choices provided. Let's analyze the options.

Let me check the options again and re-evaluate the calculation.