Let $BCDK$ be a convex quadrilateral such that $BC=BK$ and $DC=DK$. $A$ and $E$ are points such that $ABCDE$ is a convex pentagon such that $AB=BC$ and $DE=DC$ and $K$ lies in the interior of the pentagon $ABCDE$. If $\angle ABC=120^{\circ}$ and $\angle CDE=60^{\circ}$ and $BD=2$ then determine area of the pentagon $ABCDE$.
1. **Given Conditions and Notations:**
   - Let \( BC = BK = y \) and \( DC = DK = x \).
   - \( AB = BC = y \) and \( DE = DC = x \).
   - \( \angle ABC = 120^\circ \) and \( \angle CDE = 60^\circ \).
   - \( BD = 2 \).

2. **Applying the Cosine Rule in \(\triangle BCD\):**
   - We need to find the relationship between \( x \) and \( y \).
   - Using the cosine rule in \(\triangle BCD\):
     \[
     BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos(\angle BCD)
     \]
   - Since \( BC = y \) and \( CD = x \), and \(\angle BCD = 90^\circ + \alpha\) (where \(\alpha = \angle ACE\)):
     \[
     2^2 = y^2 + x^2 - 2 \cdot y \cdot x \cdot \cos(90^\circ + \alpha)
     \]
   - \(\cos(90^\circ + \alpha) = -\sin(\alpha)\):
     \[
     4 = y^2 + x^2 + 2xy \sin(\alpha)
     \]

3. **Applying the Cosine Rule in \(\triangle ABC\):**
   - Using the cosine rule in \(\triangle ABC\):
     \[
     AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)
     \]
   - Since \( AB = BC = y \) and \(\angle ABC = 120^\circ\):
     \[
     AC^2 = y^2 + y^2 - 2 \cdot y \cdot y \cdot \cos(120^\circ)
     \]
   - \(\cos(120^\circ) = -\frac{1}{2}\):
     \[
     AC^2 = y^2 + y^2 + y^2 = 3y^2
     \]
     \[
     AC = \sqrt{3}y
     \]

4. **Calculating the Area of Pentagon \(ABCDE\):**
   - The area of the pentagon \(ABCDE\) can be divided into three triangles: \(\triangle ABC\), \(\triangle ACE\), and \(\triangle ECD\).
   - Area of \(\triangle ABC\):
     \[
     [ABC] = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC) = \frac{1}{2} \cdot y \cdot y \cdot \sin(120^\circ) = \frac{1}{2} \cdot y^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} y^2
     \]
   - Area of \(\triangle ECD\):
     \[
     [ECD] = \frac{1}{2} \cdot DC \cdot DE \cdot \sin(\angle CDE) = \frac{1}{2} \cdot x \cdot x \cdot \sin(60^\circ) = \frac{1}{2} \cdot x^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} x^2
     \]
   - Area of \(\triangle ACE\):
     \[
     [ACE] = \frac{1}{2} \cdot AC \cdot CE \cdot \sin(\angle ACE) = \frac{1}{2} \cdot \sqrt{3}y \cdot x \cdot \sin(\alpha) = \frac{\sqrt{3}}{2} xy \sin(\alpha)
     \]

5. **Summing the Areas:**
   - The total area of the pentagon \(ABCDE\) is:
     \[
     [ABCDE] = [ABC] + [ACE] + [ECD]
     \]
     \[
     [ABCDE] = \frac{\sqrt{3}}{4} y^2 + \frac{\sqrt{3}}{4} x^2 + \frac{\sqrt{3}}{2} xy \sin(\alpha)
     \]
   - Using the equation \( 4 = y^2 + x^2 + 2xy \sin(\alpha) \):
     \[
     [ABCDE] = \frac{\sqrt{3}}{4} (y^2 + x^2 + 2xy \sin(\alpha)) = \frac{\sqrt{3}}{4} \cdot 4 = \sqrt{3}
     \]

The final answer is \(\boxed{\sqrt{3}}\)