In $\triangle ADB,\angle ADB=(180-18-30)^\circ=132^\circ$
Applying sine law in $\triangle ADB,$ $$\frac{AB}{\sin 132^\circ}=\frac{AD}{\sin30^\circ}\implies AD=\frac{AB}{2\sin48^\circ}$$ as $\sin132^\circ=\sin(180-132)^\circ=\sin48^\circ$
$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$
Applying sine law in $\triangle ABC,$ $$\frac{AC}{\sin 42^\circ}=\frac{AB}{\sin 96^\circ}\implies \frac{AC}{AB}=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\cos 48^\circ}{2\sin 48^\circ \cos 48^\circ}$$ (applying $\sin 2A=2\sin A\cos A$)
So, $$AC=\frac{AB}{2\sin48^\circ} \implies AC=AD$$
So, $\angle ACD=\angle ADC=\frac{(180-24)^\circ}2=78^\circ$
The first condition is already fairly restrictive. Since $\angle A = 60^\circ$, $\cos \angle A = \frac12$, so we have $$a^2 = b^2 + c^2 - bc$$ by the Law of Cosines, which is nontrivial to satisfy over the integers. (We take $a,b,c$ to be the lengths of the sides opposite $A,B,C$.)
The existence of point $D$ gives us another Diophantine constraint. Draw the altitude $BH$ from $B$ onto $AC$:
Since $\triangle BCD$ is isosceles, $H$ is the midpoint of $CD$, so $CH = 1$; by looking at the right triangle $\triangle BHC$, we get $\cos \angle C = \frac{CH}{BC} = \frac1a$. The Law of Cosines tells us that $$c^2 = a^2 + b^2 - 2b.$$
Substituting the first equation into the second yields $$c^2 = (b^2 + c^2 - bc) + b^2 - 2b \implies bc = 2b^2 - 2b \implies c=2(b-1),$$ and substituting this value of $c$ back into the first equation tells us that $$a^2 = b^2 + (2b-2)^2 - b(2b-2) \implies a^2 - 3(b-1)^2 = 1,$$ which is a Pell equation. The simplest solution to $x^2-3y^2=1$ is $x=2,y=1$, and all other solutions are obtained by taking the coefficients of $1$ and $\sqrt3$ in $(2 + \sqrt3)^k$ for some $k$.
If we use $x=2$ and $y=1$, then this gives us $a=b=c=2$, which is certainly a solution to the problem: in this case, $\triangle ABC$ is equilateral and $D=A$. This triangle has area $\sqrt 3$.
If you don't like this solution, because it's a somewhat degenerate case, then $(2+\sqrt3)^2 = 7+4\sqrt3$ yields $x=7$ and $y=4$ as the next solution. In this case, $a=7$, $b=5$, and $c=8$, which is the triangle I used for the diagram. By Heron's formula, $\triangle ABC$ has area $10 \sqrt 3$.
One of these, depending on whether you find the equilateral solution acceptable or not, is the solution with the smallest area, but we can find infinitely many triangles with integer sides that satisfy the conditions, by taking higher powers of $2+\sqrt 3$ to solve Pell's equation.
Best Answer
You can make the following.
By the law of sines get $AB$ and $AQ$, which gives $AP$ and $AQ$ and by the law of cosines get $PQ$ and $PN$.
Now, by law of cosines get $\measuredangle APQ$ and from this get $\measuredangle BPN.$
Also, by law of cosines get $BN$, by law of cosines again get $\measuredangle BPN$, which gives $\measuredangle NBM.$
Now, in $\Delta NBM$ we have now: $BM$, $BN$ and $\measuredangle NBM.$
Can you end it now?