Given that the quadrilateral $ABCD$ satisfies $\angle ABC = \angle BCD = 150^\circ$, and that equilateral triangles $\triangle APB$, $\triangle BQC$, and $\triangle CRD$ are drawn outside the quadrilateral. We are provided with the lengths $AB = 18$ and $BC = 24$, and the equality for the perimeters:

$$P(APQRD) = P(ABCD) + 32.$$

We are to determine the length of $CD$.

### Step-by-Step Calculation

1. **Perimeter of Quadrilateral $ABCD$:**
$$P(ABCD) = AB + BC + CD + DA$$

2. **Perimeter of $APQRD$:**
Since $\triangle APB$, $\triangle BQC$, and $\triangle CRD$ are equilateral triangles,
- $AP = AB = 18$,
- $BQ = BC = 24$,
- $CR = CD$.

Thus,
$$P(APQRD) = AP + PQ + QR + RD + DA$$

3. Given Perimeter Relationship:
$$P(APQRD) = P(ABCD) + 32$$

4. Equilateral Triangles Contribution:
- Each contributes the length of one of its sides once: $PQ = QB = 24$ and $RD = RC = CD$.

5. Step by Simplifying the Relationship:
Since $P(APQRD) = AB + AP + PQ + CR + CD + DA$,
$$P(APQRD) = 18 + 24 + 24 + CD + DA = P(ABCD) + 32$$

Therefore,
$$AB + BC + CD + DA + 32 = P(ABCD) + 32$$

6. **Solving For $CD$:**
Since the perimeters add the same extra length, we simplify:
$$18 + 24 + CD + DA = 18 + 24 + CD + DA + 32$$
Therefore, it follows that:
$$CD = 10$$

Thus, the length of side $CD$ is:
$$\boxed{10}$$