1. Set up the cyclic quadrilateral

Let the cyclic quadrilateral be $ABCD$ with adjacent sides $AB=2,\, BC=5$ and $\angle ABC=60^\circ.$

Since the quadrilateral is cyclic, opposite angles are supplementary. Hence $\angle ADC = 180^\circ-60^\circ=120^\circ.$

2. Use area formula by splitting into two triangles

Diagonal $AC$ divides the quadrilateral into triangles $ABC$ and $ADC$.

So, $\text{Area of quadrilateral} = [ABC]+[ADC].$

Now, $[ABC]=\frac12(AB)(BC)\sin 60^\circ = \frac12\cdot 2\cdot 5\cdot \frac{\sqrt3}{2}=\frac{5\sqrt3}{2}.$

Let $AD=x,\quad CD=y.$ Then $[ADC]=\frac12(x)(y)\sin 120^\circ = \frac12xy\cdot \frac{\sqrt3}{2}=\frac{xy\sqrt3}{4}.$

Given total area is $4\sqrt3$, so $\frac{5\sqrt3}{2}+\frac{xy\sqrt3}{4}=4\sqrt3.$

Divide by $\sqrt3$: $\frac52+\frac{xy}{4}=4.$ Thus, $\frac{xy}{4}=\frac32 \implies xy=6.$

3. Use equal diagonal expression from the two triangles

The diagonal $AC$ is common.

From triangle $ABC$, by cosine rule: $AC^2 = AB^2+BC^2-2(AB)(BC)\cos 60^\circ$ $=2^2+5^2-2\cdot 2\cdot 5\cdot \frac12=4+25-10=19.$

From triangle $ADC$, since $\angle ADC=120^\circ$, $AC^2 = x^2+y^2-2xy\cos 120^\circ.$ Now $\cos 120^\circ=-\frac12$, so $AC^2=x^2+y^2+xy.$ Hence, $x^2+y^2+xy=19.$

Using $xy=6$, $x^2+y^2=13.$

Now, $(x+y)^2=x^2+y^2+2xy=13+12=25,$ so $x+y=5$ (since lengths are positive).

4. Find the perimeter

Perimeter $=AB+BC+CD+DA=2+5+y+x=7+(x+y)=7+5=12.$

5. Check options

The perimeter is $\boxed{12}.$ So the correct option is C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.