Given: $ar(ACE) = 4, ar(CED) = 3$

$ar(ACD) = ar(CED) + ar(DEB) = 4 + 3 = 7$

Dropping a perpendicular from point $D$ on base $CD$, taking the point of intersection as $X$,
$ar(CED) = \frac{1}{2} * EX * CD$......$(2)$
Also,
$ar(ACD) = \frac{1}{2} * AC * CD$......$(3)$
Dividing $(2)$ by $(3)$,
$\frac{EX}{AC} = \frac{3}{7}$.....\frac{(4)[}{fraction]

Now, to find the area of triangle $AEB$, we drop a perpendicular from point $E$ to base $AB$, taking the point of intersection as $Y$,
$ar(AEB) = [fraction]1/2}AB * EY$
$AC = EX + EY$
Thus, $EY = AC - EX = AC - \frac{3}{7}AC$.....(from equation $4$)
$EY = \frac{4}{7}AC$

$ar(AEB) = \frac{1}{2} * AB * \frac{4}{7}AC$
$= \frac{4}{7}ar(ABC)$ ....... (since $ar(ABC) = \frac{1}{2} AB * AC$)

$ar(ABC) = ar(AEC) + ar(AEB) = 4 + \frac{4}{7}ar(ABC)$
$\frac{3}{7}ar(ABC) = 4$
$ar(ABC) = \frac{28}{3}$

Now, triangles $ABC$ and $ADB$ have the same base $AB$ and same height $AC$, so they have the same area.
Thus, $ar(ADB) = \frac{28}{3}$