A correct problem, I believe

Let $CE$ be parallel to $AB$ , then $CE$ is perpendicular to $AD$ . Drop a perpendicular $CF$ from $C$ to the extension of $AB$ .

Then, we note that $CE = AB + BC \cos \angle CBF = 46 + 13 \cos 60^\circ = 52.5$ . Since the sum of internal angles of a quadrilateral is $180^\circ$ , we note that $\angle ADC = 60^\circ$ . Then $\tan \angle ADC = \dfrac {CE}{DE}$ $\implies DE = \dfrac {CE}{\tan 60^\circ} = \dfrac {52.5}{\sqrt 3} = \dfrac {35 \sqrt 3}2$ .

By Pythagorean theorem , we have

$\begin{aligned} BD^2 & = AB^2 + DA^2 \\ & = AB^2 + (DE+{\color{#3D99F6}EA})^2 \\ & = AB^2 + (DE+{\color{#3D99F6}CF})^2 \\ & = AB^2 + (DE+{\color{#3D99F6}BC\sin 60^\circ})^2 \\ & = 46^2 + \left(\dfrac {35 \sqrt 3}2 +{\color{#3D99F6}\dfrac {13 \sqrt 3}2} \right)^2 \\ & = 2116 + 1728 = 3844 \\ \implies BD & = \sqrt{3844} = \boxed{62} \end{aligned}$

Law of cosines in $\triangle ABC$ provides $AC=\sqrt{46^2+13^2-2\times46\times13\times\cos120^\circ}=31\sqrt3$

The quadrilateral is cyclic with $BC$ a diameter of the circumcircle of $ABC$

The formula for radius of circumcircle is $r=\dfrac{abc}{4A}$ , where $a, b, c$ are sides of the triangle and $A$ is its area.

$A=\frac12 AB\times BC\times \sin 120^\circ$

$r=\dfrac{46\times13\times31\sqrt3}{4\times\frac12\times46\times13\times\frac{\sqrt3}2}=31$

$DB=2r=\boxed{62}$