Let $\mathcal S$ be the set of cool polynomials with degree $\leq 2023$ with coefficients in $[0, n]$. We aim to show that $|\mathcal S|$ is even. First, define a function $\tau$ on $\mathcal S$ as follows: Suppose that $P \in \mathcal S$ which can be written as $P(x) = x^{\theta}Q(x)$ for some polynomial $Q$ with $Q(0) \ne 0$. Observe that this is possible for each $P$ since $0 \not \in \mathcal S$. Define $\tau(P) = x^{\deg P} Q(1/x)$. Claim. $\tau$ is an involution on $\mathcal S$. Proof: To show this, we need only show that $\tau(P) \in \mathcal S$ for each $P \in \mathcal S$. Remark that the coefficients are still in $[0, n]$ and the degree is still at most $2023$, so we only need to show that $\tau(P)$ is cool. To see this, we observe that \[\prod \limits_{2 \leq x \leq p-2} x^{\deg P} P(1/x) \equiv (2 \cdot 3 \dots (p-2))^{\deg P} \prod \limits_{2 \leq x \leq p-2} P(1/x) \equiv \prod \limits_{2 \leq x \leq p-2} P(1/x) \equiv \prod \limits_{2 \leq x \leq p-2} P(x) \equiv 1 \pmod p\]where the last congruence is due to $P$ being cool (also remember that $x \mapsto 1/x$ permutes $\{2, 3, \dots, p-2\}$ modulo $p$). Thus, it suffices to show that the number of fixed points of $\tau$ are even in number. Let the set of fixed points be $\mathcal F$. Define an equivalence class $\sim$ over $\mathcal F$ where $P \sim Q$ if and only if there exists some integer (possibly negative) $m$ such that $P(x) \equiv x^m Q(x)$. It's easy to see that $\sim$ is indeed an equivalence relation. The next claim is the finish: Claim: Each equivalence class of $\mathcal F$ defined by $\sim$ has even size. Proof: First, observe that each equivalence class looks like $\{Q, Qx, \dots, Qx^{m}\}$ where $Q \in \mathcal F$ with $Q(0) \ne 0$ and $m = 2023-\deg Q$. It follows that the size of this equivalence class is $2024-m$. Therefore, to show that this set has even size, it suffices to show that each $Q \in \mathcal F$ with $Q(0) \ne 0$ satisfies $\deg Q$ even. Let $Q$ be such a polynomial and suppose for the sake of contradiction that $\deg Q \equiv 1 \pmod 2$. Since $Q$ is also a reciprocal polynomial, it follows that \[Q(x) = (x-\varepsilon)x^{\deg R} R\left(x + \frac{1}{x}\right)\]where $\varepsilon \in \{-1, 1\}$ and $R$ is some polynomial in $\mathbb{Z}[x]$. Since $Q$ is cool: \[1 \equiv \prod \limits_{2 \leq k \leq p-2} Q(k) \equiv \prod \limits_{2 \leq k \leq p-2} (k-\varepsilon) \left( \prod \limits_{2 \leq k \leq p-2} R\left(k + \frac{1}{k}\right) \right)(2 \cdot 3 \dots (p-2))^{\deg R} \pmod p \]Observe that since $\varepsilon \in \{-1, 1\}$ the first product above is exactly $-1/2$. Thus we have \[\prod_{2 \leq k \leq p-2} R\left(k + \frac{1}{k}\right) \equiv -2 \pmod p\]The time has come to put the final nail in the coffin. Observe that the numbers $k+\frac{1}{k} \mod p$ occur in pairs, once due to $k$ and $1/k$ (observe that $k = 1, -1$ are not in the product). It follows that the above product is a QR modulo $p$. But since $p \equiv 5 \pmod 8$, we know that $\left(\frac{-2}{p}\right) = -1$, a contradiction. It follows that each such $Q$ has even degree, and the claim follows readily now. Since each equivalence class of $\mathcal F$ has even size, so does $\mathcal F$ and so does $\mathcal S$, as required.

To prove that the number of cool polynomials is even, we will find a pairing of them and prove that no cool polynomial is paired to itself. Lemma: If $Q$ is cool, so is $x^{2023}Q(\frac{1}{x})$. Proof: Assume that $Q$ is cool. Denote $x^{2023}Q(\frac{1}{x})$ by $P(x)$. First Notice that $\operatorname{deg}P \leq 2023$ and all of its coefficients are also less than or equal to $n$. We have: $$P(2)\cdot P(3) \cdot \ldots \cdot P(p-2)-1 \equiv 2^{2023}Q(2)\cdot 3^{2023}Q(3) \cdot \ldots \cdot (p-2)^{2023}Q(p-2)-1 \equiv ((p-2)!)^{2023} \cdot Q(2) \cdot Q(3) \cdot \ldots \cdot Q(p-2)-1 \equiv 0 \pmod p$$Therefore, $P(x)=x^{2023}Q(\frac{1}{x})$ is also cool. Notice that $Q(x) \to x^{2023}Q(\frac{1}{x})$ is an involution, so we only need to prove it has no fixed points. Assume that a cool polynomial $Q$ satisfies $Q(x)= x^{2023}Q(\frac{1}{x})$. Pair the residues $2,3,\ldots,p-2 \pmod p$ to pairs of $a,\frac{1}{a} \pmod p$ and denote the set of pairs by $S$. Then: $$1 \equiv Q(2)\cdot Q(3) \cdot \ldots \cdot Q(p-2) \equiv \prod_{(a,\frac{1}{a}) \in S} Q(a)Q(\frac{1}{a}) \equiv \prod_{(a,\frac{1}{a}) \in S} a^{2023}Q(\frac{1}{a})^2$$Therefore, $ \prod_{(a,\frac{1}{a}) \in S} a^{2023}$ is a quadric residue $\pmod p$, which means that $ \prod_{(a,\frac{1}{a}) \in S} a$ is a quadric residue $\pmod p$, but notice that the number of quadratic nonresidues in $2,3, \ldots p-2$ is $\frac{p-1}{2}$ (because the number of quadratic nonresidues modulo $p$ is $\frac{p-1}{2}$ and all of the residues we omitted are quadric residues), and $a$ is a quadric residue if and only if $\frac{1}{a}$ is a quadric residue, so the number of quadric nonresidues in the product $ \prod_{(a,\frac{1}{a}) \in S} a$ is $\frac{p-1}{4} = 2k+1$ which is odd, i.e. this product is not a quadric residue, which is a contradiction. Therefore, $Q(x) \to x^{2023}Q(\frac{1}{x})$ is an involution on the cool polynomials without any fixed points, which means the number of cool polynomials is even.

Obvious.

FTSOC, assume that there is a fixed point in this pairing, it suffices to prove that "palindromic" polynomials can not satisfy the condition. Now pair $x$ with $x^{-1}$ (modulo $p$) then $$\prod_{2}^{p-2} Q(2) \cdot Q(3) \cdots Q(p-2) \equiv \prod_{a \in {2,3, \cdots \frac{p-1}{2}}} a^{2023}Q\left(\frac{1}{a}\right)^2 $$and hence we have $\prod a$ is a QR, now the number of NQR in ${1,2, \cdots \frac{p-1}{2}}$ is odd, contradiction.

Obvious.

FTSOC, assume that there is a fixed point in this pairing, it suffices to prove that "palindromic" polynomials can not satisfy the condition. Now pair $x$ with $x^{-1}$ (modulo $p$) then $$\prod_{2}^{p-2} Q(2) \cdot Q(3) \cdots Q(p-2) \equiv \prod_{a \in {2,3, \cdots \frac{p-1}{2}}} a^{2023}Q\left(\frac{1}{a}\right)^2 $$and hence we have $\prod a$ is a QR, now the number of NQR in ${1,2, \cdots \frac{p-1}{2}}$ is odd, contradiction.

Obvious.

FTSOC, assume that there is a fixed point in this pairing, it suffices to prove that "palindromic" polynomials can not satisfy the condition. Now pair $x$ with $x^{-1}$ (modulo $p$) then $$\prod_{2}^{p-2} Q(2) \cdot Q(3) \cdots Q(p-2) \equiv \prod_{a \in {2,3, \cdots \frac{p-1}{2}}} a^{2023}Q\left(\frac{1}{a}\right)^2 $$and hence we have $\prod a$ is a QR, now the number of NQR in ${1,2, \cdots \frac{p-1}{2}}$ is odd, contradiction.