1. **Define the function $d(a, b)$:**
Given a prime $p$, let $d(a, b)$ be the number of integers $c$ such that $1 \leq c < p$, and the remainders when $ac$ and $bc$ are divided by $p$ are both at most $\frac{p}{3}$.

2. Express the given expression in terms of vectors:
Let $v_a$ be a vector in $\mathbb{R}^{p-1}$ such that the $c$-th term of $v_a$ is:
$$v_a(c) = \begin{cases} x_a + 1 & \text{if the remainder of } ac \text{ is at most } \frac{p}{3}, \\ 0 & \text{otherwise}. \end{cases}$$
Similarly, let $u_a$ be a vector in $\mathbb{R}^{p-1}$ such that the $c$-th term of $u_a$ is:
$$u_a(c) = \begin{cases} x_a & \text{if the remainder of } ac \text{ is at most } \frac{p}{3}, \\ 0 & \text{otherwise}. \end{cases}$$

3. Rewrite the sums using these vectors:
The given expression can be rewritten as:
$$\sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1} d(a, b) (x_a + 1)(x_b + 1)} = \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle v_a, v_b \rangle}$$
and
$$\sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1} d(a, b) x_a x_b} = \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle u_a, u_b \rangle}.$$

4. Apply the triangle inequality:
By the triangle inequality for vectors, we have:
$$\| v_a \| \leq \| u_a \| + \| \mathbf{1} \|,$$
where $\mathbf{1}$ is the vector with all entries equal to 1. Since $\| \mathbf{1} \| = \sqrt{p-1}$, we get:
$$\sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle v_a, v_b \rangle} \leq \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle u_a, u_b \rangle} + \sqrt{p-1}.$$

5. Determine the maximum value:
The maximum value of the given expression is achieved when the difference between the two square roots is maximized. This occurs when:
$$\sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle v_a, v_b \rangle} - \sqrt{\sum_{a=1}^{p-1} \sum_{b=1}^{p-1} \langle u_a, u_b \rangle} = \sqrt{p-1}.$$

The final answer is $\boxed{\sqrt{p-1}}$.