We first investigate what primes divide $d$. Notice that a prime $p$ divides $P(n)$ for all $n \geq 2024$ if and only if $\left\{1^{3}, 2^{3}, \ldots, 40^{3}\right\}$ contains all residues in modulo $p$. Hence, $p \leq 40$. Moreover, $x^{3} \equiv 1$ must not have other solution in modulo $p$ than 1, so $p \not \equiv 1(\bmod 3)$. Thus, the set of prime divisors of $d$ is $S=\{2,3,5,11,17,23,29\}$. Next, the main claim is that for all prime $p \in S$, the minimum value of $\nu_{p}(P(n))$ across all $n \geq 2024$ is $\left\lfloor\frac{40}{p}\right\rfloor$. To see why, note the following: - Lower Bound. Note that for all $n \in \mathbb{Z}$, one can group $n-1^{3}, n-2^{3}, \ldots, n-40^{3}$ into $\left\lfloor\frac{40}{p}\right\rfloor$ contiguous blocks of size $p$. Since $p \not \equiv 1(\bmod 3), x^{3}$ span through all residues modulo $p$, so each block will have one number divisible by $p$. Hence, among $n-1^{3}, n-2^{3}, \ldots, n-40^{3}$, at least $\left\lfloor\frac{40}{p}\right\rfloor$ are divisible by $p$, implying that $\nu_{p}(P(n))>\left\lfloor\frac{40}{p}\right\rfloor$. - Upper Bound. We pick any $n$ such that $\nu_{p}(n)=1$ so that only terms in form $n-p^{3}, n-(2 p)^{3}$, $\ldots$ are divisible by $p$. Note that these terms are not divisible by $p^{2}$ either, so in this case, we have $\nu_{p}(P(n))=\left\lfloor\frac{40}{p}\right\rfloor$. Hence, $\nu_{p}(d)=\left\lfloor\frac{40}{p}\right\rfloor$ for all prime $p \in S$. Thus, the answer is $$\sum_{p \in S}\left\lfloor\frac{40}{p}\right\rfloor=\left\lfloor\frac{40}{2}\right\rfloor+\left\lfloor\frac{40}{3}\right\rfloor+\left\lfloor\frac{40}{5}\right\rfloor+\left\lfloor\frac{40}{11}\right\rfloor+\left\lfloor\frac{40}{17}\right\rfloor+\left\lfloor\frac{40}{23}\right\rfloor+\left\lfloor\frac{40}{29}\right\rfloor=48$$