If $\mathcal{L}$ is such that $K_X=\mathcal{L}^{\otimes b}$, then $$P'(t+b)=\chi(X, \, (\mathcal{L}^{\vee})^{\otimes t+b}) = \chi(X, K_X \otimes \mathcal{L}^{t+b})=P(t)$$
Francesco Polizzi
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