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Include a comment by René
jmc
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The only natural thing I can think of is the following:

Let $R_{\mathcal{P}}$ be the ring of integers of $K_{\mathcal{P}}$ (the completion at the prime). Let $S$ be the spectrum of $R_{\mathcal{P}}$. Since $A$ has good reduction, there is an abelian scheme (the Néron model) $\mathcal{A}$ over $S$, with $A$ as generic fibre, and the reduction as special fibre (which we will denote $A_{0}$.

Now look at maps $S[\varepsilon] \to \mathcal{A}$ that restrict to the identity section $e \colon S \to \mathcal{A}$, where $S[\varepsilon]$ is $S \times \mathrm{Spec}(\mathbb{Z}[\varepsilon]/(\varepsilon^{2}))$. Such maps $S[\varepsilon] \to \mathcal{A}$ restrict to tangent vectors on the generic fibre and the special fibre. So you get maps $$ T_{e}A \longleftarrow \mathrm{Hom}_{S}(S[\varepsilon],\mathcal{A}) \longrightarrow T_{e}A_{0}. $$


Like René observed below in the comments:

the middle term is a tangent space in its own right (and a free $R_{\mathcal{P}}$-module of rank $\dim A$), which naturally embeds into $T_{e}A$ and has a reduction map to $T_{e}A_{0}$.

jmc
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