This is a K-Truss. I will try to provide a solution.
Corrections: There should be a cyan member in the truss, otherwise, it is unstable.
As shown above, the system will be cut by curved line (as highlighted by the red line). This is a special case to the method of sections wherein the rule allows cutting of more than 3 members provided that all member except one are collinear at a point. Separating the left side of the system gives:
The reactions have been solved by static equations. In the above image, we only need the horizontal unknowns $LK$ and $CD$, thus:
Summing up moments at C yields:
$$15\ (8)=LK\ (5)$$
$$LK\ =\ 24\ kN\ (Tension)$$
I will leave the sign as is so as to simplify the analysis of sections.
Next is summing up moments at L yields:
$$15\ (8)\ -\ CD\ (5) = 0$$
$$CD\ =\ -24\ kN$$
Note that we do not need to solve the values of member BP and MP as they are not needed in the solution.
Next, we cut another section from the system as shown below:
and separating the left section:
Note that members $CD$ and $LK$ has been solved as denoted in yellow color. Two equations and two unknowns will be needed, thus, summing up forces along x-axis and y-axis yields the following equations respectively:
$$QD\ [4/sqrt(22.25)]\ +\ QK\ [4/sqrt(22.25)]\ +\ CD\ +\ LK\ =\ 0$$
$$QD\ [2.5/sqrt(22.25_]\ -\ QK\ [2.5/sqrt(22.25)]\ -\ 5\ +\ 15\ =\ 0$$
$$QD\ =\ -9.434\ kN$$
$$QK\ =\ +9.434\ kN$$
Note that we cannot still use method of joints at joint $D$, therefore, we will continue to solve for more members of the system as:
Separating the right sections:
Summing up moments at J yields:
$$20\ (8)\ +\ ED\ (5)\ =\ 0$$
$$ED\ = \ -32\ kN$$
Now we can use method of joints at joint $D$ because there are only two unknowns left as:
Summing up forces along x-axis yields:
$$-DC\ -\ DQ\ [4/sqrt(22.25)]\ +\ DE\ +\ DR\ [(4/sqrt(22.25)]\ = \ 0$$
$$DR\ = \ 0$$
Summing up forces in the y-axis yields:
$$-DQ\ [2.5/sqrt(22.25)]\ +\ -DK\ =\ 0$$
$$DK\ =\ 5\ kN$$