To solve the problem, we need to find the probability that $\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}$ for $x$ chosen randomly from the interval $5 \leq x \leq 15$. We are given $P(x) = x^2 - 3x - 9$.

1. Substitution and Simplification:
Let's make the substitution $y = 2x - 3$. Then, we have:
$$P(x) = \frac{y^2 - 45}{4}$$
We need to find the solutions to:
$$\left\lfloor \sqrt{\frac{y^2 - 45}{4}} \right\rfloor = \sqrt{\frac{\lfloor y \rfloor^2 - 45}{4}}$$
with the new bounds $y \in [7, 27]$.

2. Perfect Square Condition:
Since the left side is an integer, $\frac{\lfloor y \rfloor^2 - 45}{4}$ must be a perfect square. Let $\lfloor y \rfloor = a$, then:
$$a^2 - 45 = 4b^2 \implies (a - 2b)(a + 2b) = 45$$
We need to find pairs $(a - 2b, a + 2b)$ that multiply to 45. The factor pairs of 45 are $(1, 45), (3, 15), (5, 9)$.

3. **Solving for $a$ and $b$**:
- For $(a - 2b, a + 2b) = (1, 45)$:
$$a - 2b = 1 \quad \text{and} \quad a + 2b = 45$$
Solving these, we get:
$$a = 23, \quad b = 11$$
- For $(a - 2b, a + 2b) = (3, 15)$:
$$a - 2b = 3 \quad \text{and} \quad a + 2b = 15$$
Solving these, we get:
$$a = 9, \quad b = 3$$
- For $(a - 2b, a + 2b) = (5, 9)$:
$$a - 2b = 5 \quad \text{and} \quad a + 2b = 9$$
Solving these, we get:
$$a = 7, \quad b = 1$$

4. Finding the Intervals:
- For $a = 23$:
$$11 \leq \sqrt{\frac{y^2 - 45}{4}} < 12 \implies 23 \leq y < \sqrt{621}$$
- For $a = 9$:
$$3 \leq \sqrt{\frac{y^2 - 45}{4}} < 4 \implies 9 \leq y < \sqrt{109}$$
- For $a = 7$:
$$1 \leq \sqrt{\frac{y^2 - 45}{4}} < 2 \implies 7 \leq y < \sqrt{61}$$

5. Calculating the Total Length of Success Intervals:
The total length of the interval from which we choose $y$ is $27 - 7 = 20$. The total length of the success intervals is:
$$(\sqrt{61} - 7) + (\sqrt{109} - 9) + (\sqrt{621} - 23) = \sqrt{61} + \sqrt{109} + \sqrt{621} - 39$$

6. Probability Calculation:
The probability is:
$$\frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}$$

7. Identifying Constants:
Comparing with $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$, we have:
$$a = 61, \quad b = 109, \quad c = 621, \quad d = 39, \quad e = 20$$

8. Summing the Constants:
$$a + b + c + d + e = 61 + 109 + 621 + 39 + 20 = 850$$

The final answer is $\boxed{850}$.