Jimmy invites Kima, Lester, Marlo, Namond, and Omar to dinner. There are nine chairs at Jimmy's round dinner table. Jimmy sits in the chair nearest the kitchen. How many different ways can Jimmy's five dinner guests arrange themselves in the remaining $8$ chairs at the table if Kima and Marlo refuse to be seated in adjacent chairs?
1. **Fix Jimmy's Position**: Since Jimmy is fixed in the chair nearest the kitchen, we have 8 remaining chairs for the 5 guests.

2. **Total Arrangements Without Restrictions**: The total number of ways to arrange 5 guests in 8 chairs is given by:
   \[
   P(8, 5) = 8 \times 7 \times 6 \times 5 \times 4 = 6720
   \]

3. **Counting the Unwanted Arrangements (Kima and Marlo Adjacent)**:
   - Treat Kima and Marlo as a single "block" or "super guest". This reduces the problem to arranging 4 guests (the block and the other 3 guests) in 8 chairs.
   - The number of ways to place the "block" in 8 chairs is 7 (since the block can start at any of the first 7 positions, and the second part of the block will be adjacent).
   - For each of these 7 positions, the remaining 3 guests can be arranged in the remaining 6 chairs:
     \[
     P(6, 3) = 6 \times 5 \times 4 = 120
     \]
   - Since Kima and Marlo can switch places within the block, we multiply by 2:
     \[
     7 \times 120 \times 2 = 1680
     \]

4. **Complementary Counting**: Subtract the unwanted arrangements from the total arrangements:
   \[
   6720 - 1680 = 5040
   \]

Conclusion:
\[
\boxed{5040}
\]