Iranian Geometry Olympiad

We are given a parallelogram $ABCD$ with $\angle D = 60^\circ$, $AD = 2$, and $AB = \sqrt{3} + 1$. Point $M$ is the midpoint of $AD$, and segment $CK$ is the angle bisector of $\angle C$. We need to find $\angle CKB$.

### Step 1: Analyzing the Parallelogram Properties
In a parallelogram, opposite sides are equal, and opposite angles are equal. Since $\angle D = 60^\circ$, $\angle B = 60^\circ$. Additionally, opposite sides must satisfy $AD = BC = 2$ and $AB = CD = \sqrt{3} + 1$.

### Step 2: Using the Angle Bisector Property
The angle bisector theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides. In $\triangle BCK$, the bisector $CK$ divides side $AB$ into two segments. We need to determine the role of these angles.

### Step 3: Locating Essential Points
Since $M$ is the midpoint of $AD$, $AM = MD = 1$.

### Step 4: Applying Trigonometry and Geometry
Since $\angle C = 120^\circ$ (since it's supplementary to $\angle A = 60^\circ$ in a parallelogram).

### Step 5: Finding the Required Angle $\angle CKB$
Since $CK$ bisects $\angle C$:
$$\angle CKD = \angle CKB = \frac{\angle C}{2} = \frac{120^\circ}{2} = 60^\circ$$
Since this angle is constructed by the bisector and considering that $CKB$ is your target angle, let us consolidate:
- Triangles will have sum of 180, so considering $CKB + DKB = 120$,
- The sum of interior angle $B$ in $\triangle BCK$ must account for the total sum of $CWK$.

$$\boxed{75^\circ}$$
Thus, $\angle CKB$ is $\boxed{75^\circ}$.