Correct Answer: 90°

Solution : A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, and AC and BD intersect at O.
$\angle$ CAD = 46$^\circ$
ABCD is a cyclic quadrilateral,
$\angle$ A + $\angle$ C = $\angle$ B + $\angle$ D = 180$^\circ$
AD = CD
⇒ $\angle$ CAD = $\angle$ DCA = 46$^\circ$
⇒ $\angle$ ADC = 180$^\circ$ - 92$^\circ$ = 88$^\circ$
⇒ $\angle$ ABC = 180$^\circ$ - 88$^\circ$ = 92$^\circ$
AB = BC 
⇒ $\angle$ BAC = $\angle$ BCA = $\frac{180^\circ-92^\circ}{2}$=$\frac{88^\circ}{2}$ = 44$^\circ$
Now,
$\angle$ ABO = $\angle$ DCA = $\angle$ DAC = 46$^\circ$ (angles made by the same arc on the same side of the circle are equal)
⇒ $\angle$ AOB = 180$^\circ$ - (46$^\circ$ + 44$^\circ$) = 90$^\circ$
$\therefore$ $\angle$ AOB is 90$^\circ$ 
Hence, the correct answer is 90$^\circ$.