Let $A$ and $B$ be two non-empty subsets of $X = \{1, 2, . . . , 8 \}$ with $A \cup B = X$ and $A \cap B = \emptyset$. Let $P_A$ be the product of all elements of $A$ and let $P_B$ be the product of all elements of $B$. Find the minimum possible value of sum $P_A +P_B$. 

PS. It is a variation of [url=https://artofproblemsolving.com/community/c6h2267998p17621980]JBMO Shortlist 2019 A3 [/url]
To find the minimum possible value of the sum \( P_A + P_B \), where \( P_A \) is the product of all elements in subset \( A \) and \( P_B \) is the product of all elements in subset \( B \), we need to consider the constraints \( A \cup B = X \) and \( A \cap B = \emptyset \). This means \( A \) and \( B \) are disjoint and their union is the set \( X = \{1, 2, \ldots, 8\} \).

1. **Calculate the product of all elements in \( X \):**
   \[
   8! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 40320
   \]

2. **Express the problem in terms of \( P_A \) and \( P_B \):**
   Since \( A \cup B = X \) and \( A \cap B = \emptyset \), we have:
   \[
   P_A \cdot P_B = 8!
   \]
   We need to minimize \( P_A + P_B \).

3. **Use the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality):**
   \[
   \frac{P_A + P_B}{2} \geq \sqrt{P_A \cdot P_B}
   \]
   Substituting \( P_A \cdot P_B = 8! \):
   \[
   \frac{P_A + P_B}{2} \geq \sqrt{8!}
   \]
   \[
   P_A + P_B \geq 2\sqrt{8!}
   \]
   Calculate \( \sqrt{8!} \):
   \[
   \sqrt{8!} = \sqrt{40320} \approx 200.8
   \]
   Therefore:
   \[
   P_A + P_B \geq 2 \times 200.8 = 401.6
   \]

4. **Find integer values for \( P_A \) and \( P_B \):**
   Since \( P_A \) and \( P_B \) must be integers, we need to find the closest integer values that satisfy \( P_A \cdot P_B = 40320 \) and \( P_A + P_B \) is minimized.

5. **Check possible combinations:**
   We need to find subsets \( A \) and \( B \) such that \( P_A \leq \sqrt{8!} \approx 200.8 \) and \( P_B \geq \sqrt{8!} \approx 200.8 \).

   - Consider \( A = \{2, 3, 4, 8\} \):
     \[
     P_A = 2 \cdot 3 \cdot 4 \cdot 8 = 192
     \]
     Then \( B = \{1, 5, 6, 7\} \):
     \[
     P_B = 1 \cdot 5 \cdot 6 \cdot 7 = 210
     \]
     Thus:
     \[
     P_A + P_B = 192 + 210 = 402
     \]

6. **Verify if this is the minimum:**
   We need to check if there are any other combinations that yield a smaller sum. However, given the constraints and the AM-GM bound, \( 402 \) is the smallest sum achievable with the given subsets.

The final answer is \( \boxed{402} \).