For the sake of simplicity, let us assume that $\mathcal{L}$ is a polarization, and that we are computing the Hilbert polynomial of $\mathcal{L}$ with respect to itself, i.e. $P(t)=\chi(X, \, \mathcal{L}^{\otimes t})$.
If $\mathcal{L}$ is such that $K_X=(\mathcal{L^{\vee}})^{\otimes b}$, then by Serre duality we obtain $$P'(t+b)=\chi(X, \, (\mathcal{L}^{\vee})^{\otimes t+b}) = \chi(X, K_X \otimes \mathcal{L}^{t+b})=\chi(X, \, \mathcal{L}^{\otimes t})= P(t).$$$$P'(t+b)=\chi(X, \, (\mathcal{L}^{\vee})^{\otimes t+b}) = \chi(X, K_X \otimes \mathcal{L}^{\otimes t+b})=\chi(X, \, \mathcal{L}^{\otimes t})= P(t).$$
For instance, if $X=\mathbb{P}^{n-1}$ and $\mathcal{L}=\mathcal{O}_{\mathbb{P}^{n-1}}(1)$, we have $K_X = (\mathcal{L}^{\vee})^{\otimes n}$ and so $$P'(t+n)=P(t).$$