Where in the title [...] denotes the ceiling function which is used for a shorter title. This sum arises as corrections terms to the number of reducible quadratics of height $N$. Looking for simplification of the sum
$${A}_{N} = \sum_{k = 0}^{\left\lceil{p/3}\right\rceil - 2} \sum_{i = 1}^{\left\lceil{p/\left({k + 1}\right)}\right\rceil - 3} \sum_{j = 0}^{p - \left({k + 1}\right) \left({i + 2}\right) - 1} {\delta}_{N, i\, p + j + k + 1}$$
Where ${\delta}_{N,X}$ is the Kronika Delta function. Using Iverson Brackets we can write after simplification
$${A}_{N} = \sum_{k = 0}^{\left\lceil{p/3}\right\rceil - 2} \sum_{i = 1}^{\left\lceil{p/\left({k + 1}\right)}\right\rceil - 3} \left[{i\, p + k + 1 \le N \le i\, p - i\, k + p - i - k - 2}\right]$$
After rearranging this, we get
$${A}_{N} = \sum_{k = 0}^{\left\lceil{p/3}\right\rceil - 2} \sum_{i = 1}^{\left\lceil{p/\left({k + 1}\right)}\right\rceil - 3} \left[{\frac{N + k + 2 - p}{p - k - 1} \le i \le \frac{N - k - 1}{p}}\right]$$
The problem is solving for the bounds of $i$ for $k$ where we have ceiling functions involving $k$ to reduce the sum over $i$.
The methods from Find the asymptotic form as $N \rightarrow \infty$ of $\sum_{a = 1}^{N} \sum_{u = 1}^{a - 2} \sum_{v = u + 1}^{a - 1} {\delta}_{N = u\, a + v}$ were illustrative in solving similar sums.