Based on the information given in the question, a diagram is constructed where BM is the median of AC. Since the question deals with areas, we construct BM as the altitude for the triangle.$\angle\ BDA=\theta$ (given) and$\angle\ BAD=\alpha$ is assumed. The following figure is obtained.
BM is the height, and so$\angle\ BMA=\angle\ BMD=90^{\circ\ }$
So, in $\triangle\ BAM$,
$\sin\alpha\ =\frac{BM}{AB}=\ BM=c\ \sin\alpha$
$\cos\ \alpha\ =\frac{AM}{AB}=AM=c\cos\alpha$
Again, in $\triangle\ BDM$,
$\sin\theta\ =\frac{BM}{BD}=\ BM=k\ \sin\theta$
$\cos\ \theta\ =\frac{DM}{BD}=DM=k\cos\theta$
Now, area of$\triangle\ ABD$ can be calculated as$\dfrac{1}{2}\times\ AD\times\ BM=\frac{1}{2}\times\ \left(AM+BM\right)\times\ BM$
Putting the values obtained we have,
$ar\triangle\ ABD=\frac{1}{2}\times\ \left(c\ \cos\alpha\ +k\ \cos\theta\ \right)\times\ k\ \sin\theta$ ________(A)
Now,$c\ \sin a=\ k\ \sin\theta\ =\sin\ \alpha\ =\dfrac{k\sin\theta\ }{c}$
So,$\cos\ \alpha\ =\sqrt{\ 1-\left(\frac{k\sin\theta\ }{c}\ \right)^2}=\dfrac{\sqrt{\ c^2-k^2\sin^2\theta\ }}{c}$
Substituting this in (A) we have
$ar\triangle\ ABD=\frac{1}{2}\times\ \left(\sqrt{\ c^2-k^2\sin^2\theta\ }\ +k\ \cos\theta\ \right)\times\ k\ \sin\theta$
or, $ar\triangle\ ABD=\frac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +k^2\ \cos\theta\sin\theta\ \right)$
or, $ar\triangle\ ABD=\frac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\frac{k^2}{2}\times\ \ 2\cos\theta\sin\theta\ \right)$
or, $ar\triangle\ ABD=\dfrac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\dfrac{k^2}{2}\sin2\theta\ \right)$
Now, since BD is the median, this means the area of$ar\triangle\ ABC=2\ ar\triangle\ ABD$
so ,$ar\triangle\ ABC=\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\dfrac{k^2}{2}\sin2\theta$ ,
which is option B.
