Answer: Option A
:
A and C
$\angle EDC$ = $\angle ECD$ (Angles of an equilateral triangle)
And $\angle ADC$ = $\angle BCD$ (Angles of a rectangle)
$\therefore \angle EDC+\angle ADC=\angle ECD+\angle BCD$ 
$\Rightarrow \angle EDA=\angle ECB$
In $\Delta DAE$ and $\Delta CBE$,
$DE=CE$ (Sides of an equilateral triangle)
$DA=BC$ (Sides of a rectangle)
$\angle EDA$ = $\angle ECB$ (proved)
Hence, $\Delta DAE\cong \Delta CBE$ (SAS congruence)
Also, EA = EB  (C.P.C.T.C.)
$\Rightarrow \Delta EAB$ is an isosceles triangle.
It is given that, ABCD is a rectangle.
$\Rightarrow DA\ne DC$ 
Also, DEC is an equilateral triangle.
$\Rightarrow DC=DE$
$\therefore DA\ne DE$
 Hence, DAE is not an isosceles triangle.
Also, it can not be proved that $\angle DAE={15}^{\circ }$.