A cube $ABCDA'B'C'D'$ is given  with an edge of length $2$ and vertices marked as in the figure. The point $K$ is center of the edge $AB$. The plane containing the points $B',D', K$ intersects the edge $AD$ at point $L$. Calculate the volume of the  pyramid with apex $A$ and base the quadrilateral $D'B'KL$.
[img]https://cdn.artofproblemsolving.com/attachments/7/9/721989193ffd830fd7ad43bdde7e177c942c76.png[/img]
1. **Identify the coordinates of the vertices:**
   - Let the cube be positioned in a 3D coordinate system with \( A = (0, 0, 0) \), \( B = (2, 0, 0) \), \( C = (2, 2, 0) \), \( D = (0, 2, 0) \), \( A' = (0, 0, 2) \), \( B' = (2, 0, 2) \), \( C' = (2, 2, 2) \), and \( D' = (0, 2, 2) \).
   - The point \( K \) is the midpoint of \( AB \), so \( K = (1, 0, 0) \).

2. **Find the equation of the plane containing \( B' \), \( D' \), and \( K \):**
   - The coordinates of \( B' \) are \( (2, 0, 2) \), \( D' \) are \( (0, 2, 2) \), and \( K \) are \( (1, 0, 0) \).
   - The vectors \( \overrightarrow{B'D'} \) and \( \overrightarrow{B'K} \) are:
     \[
     \overrightarrow{B'D'} = (0 - 2, 2 - 0, 2 - 2) = (-2, 2, 0)
     \]
     \[
     \overrightarrow{B'K} = (1 - 2, 0 - 0, 0 - 2) = (-1, 0, -2)
     \]
   - The normal vector \( \mathbf{n} \) to the plane is given by the cross product \( \overrightarrow{B'D'} \times \overrightarrow{B'K} \):
     \[
     \mathbf{n} = \begin{vmatrix}
     \mathbf{i} & \mathbf{j} & \mathbf{k} \\
     -2 & 2 & 0 \\
     -1 & 0 & -2
     \end{vmatrix} = \mathbf{i}(2 \cdot -2 - 0 \cdot 0) - \mathbf{j}(-2 \cdot -2 - 0 \cdot -1) + \mathbf{k}(-2 \cdot 0 - 2 \cdot -1)
     \]
     \[
     \mathbf{n} = \mathbf{i}(-4) - \mathbf{j}(4) + \mathbf{k}(2) = (-4, -4, 2)
     \]
   - The equation of the plane is:
     \[
     -4(x - 2) - 4(y - 0) + 2(z - 2) = 0 \implies -4x + 8 - 4y + 2z - 4 = 0 \implies -4x - 4y + 2z + 4 = 0 \implies 2x + 2y - z = 2
     \]

3. **Find the intersection point \( L \) of the plane with the edge \( AD \):**
   - The edge \( AD \) has parametric form \( (0, 2t, 0) \) where \( t \in [0, 1] \).
   - Substitute \( (0, 2t, 0) \) into the plane equation \( 2x + 2y - z = 2 \):
     \[
     2(0) + 2(2t) - 0 = 2 \implies 4t = 2 \implies t = \frac{1}{2}
     \]
   - Thus, \( L = (0, 1, 0) \).

4. **Calculate the volume of the pyramid \( A \) with base \( D'B'KL \):**
   - The vertices of the pyramid are \( A = (0, 0, 0) \), \( D' = (0, 2, 2) \), \( B' = (2, 0, 2) \), \( K = (1, 0, 0) \), and \( L = (0, 1, 0) \).
   - The volume \( V \) of a pyramid with base area \( B \) and height \( h \) is given by:
     \[
     V = \frac{1}{3} \times B \times h
     \]
   - The base \( D'B'KL \) is a quadrilateral, and we can split it into two triangles \( D'B'K \) and \( D'KL \).

5. **Calculate the area of the triangles \( D'B'K \) and \( D'KL \):**
   - For \( \triangle D'B'K \):
     \[
     \text{Area} = \frac{1}{2} \left| \overrightarrow{D'B'} \times \overrightarrow{D'K} \right|
     \]
     \[
     \overrightarrow{D'B'} = (2, -2, 0), \quad \overrightarrow{D'K} = (1, -2, -2)
     \]
     \[
     \overrightarrow{D'B'} \times \overrightarrow{D'K} = \begin{vmatrix}
     \mathbf{i} & \mathbf{j} & \mathbf{k} \\
     2 & -2 & 0 \\
     1 & -2 & -2
     \end{vmatrix} = \mathbf{i}(4 - 0) - \mathbf{j}(-4 - 0) + \mathbf{k}(-4 - (-2)) = (4, 4, -2)
     \]
     \[
     \left| \overrightarrow{D'B'} \times \overrightarrow{D'K} \right| = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6
     \]
     \[
     \text{Area} = \frac{1}{2} \times 6 = 3
     \]
   - For \( \triangle D'KL \):
     \[
     \text{Area} = \frac{1}{2} \left| \overrightarrow{D'K} \times \overrightarrow{D'L} \right|
     \]
     \[
     \overrightarrow{D'K} = (1, -2, -2), \quad \overrightarrow{D'L} = (0, -1, -2)
     \]
     \[
     \overrightarrow{D'K} \times \overrightarrow{D'L} = \begin{vmatrix}
     \mathbf{i} & \mathbf{j} & \mathbf{k} \\
     1 & -2 & -2 \\
     0 & -1 & -2
     \end{vmatrix} = \mathbf{i}(4 - 2) - \mathbf{j}(-2 - 0) + \mathbf{k}(-1 - 0) = (2, 2, -1)
     \]
     \[
     \left| \overrightarrow{D'K} \times \overrightarrow{D'L} \right| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3
     \]
     \[
     \text{Area} = \frac{1}{2} \times 3 = 1.5
     \]
   - The total area of the base \( D'B'KL \) is:
     \[
     3 + 1.5 = 4.5
     \]

6. **Calculate the height of the pyramid:**
   - The height is the perpendicular distance from \( A \) to the plane \( 2x + 2y - z = 2 \).
   - The distance \( d \) from a point \( (x_0, y_0, z_0) \) to a plane \( ax + by + cz + d = 0 \) is given by:
     \[
     d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}
     \]
   - For \( A = (0, 0, 0) \) and the plane \( 2x + 2y - z - 2 = 0 \):
     \[
     d = \frac{|2(0) + 2(0) - 0 - 2|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{|-2|}{\sqrt{4 + 4 + 1}} = \frac{2}{\sqrt{9}} = \frac{2}{3}
     \]

7. **Calculate the volume of the pyramid:**
   \[
   V = \frac{1}{3} \times 4.5 \times \frac{2}{3} = \frac{1}{3} \times 4.5 \times \frac{2}{3} = \frac{9}{9} = 1
   \]

The final answer is \(\boxed{1}\).