= Solution
Let $K^{\mathrm{nr}}$ be the maximal <unramified extension> of $K$. Because $p\nmid n$, multiplication by $n$ on the special fibre is a separable isogeny and is surjective on $\widetilde E(\overline k)$. Choose a point $\widetilde Q$ with $[n]\widetilde Q=\widetilde P$ and lift it, after a finite unramified extension, to $Q_0$. Then $R=P-[n]Q_0$ lies in the kernel of reduction.
On the <formal group of an elliptic curve>, multiplication by $n$ has the form
$$
[n]_F(T)=nT+O(T^2).
$$
Since $n$ is a unit in $\mathcal O_K$, the <invertible morphism criterion for formal group laws> makes $[n]_F$ an automorphism of $\pi\mathcal O_{K^{\mathrm{nr}}}$. Hence $R=[n]S$ for a unique point $S$ in the kernel of reduction, and $Q=Q_0+S$ satisfies $[n]Q=P$. Moreover, good reduction makes the finite group scheme $E[n]$ étale over $\mathcal O_K$, so all its points are defined over an unramified extension. Every point of $[n]^{-1}P=Q+E[n]$ is therefore unramified, proving that $K([n]^{-1}P)/K$ is unramified.
Multiplication by $n$ is already an automorphism of the formal kernel, so the exact reduction sequence shows that $E(K)/nE(K)$ is finite. Choose representatives $P_1,\ldots,P_s$. Each becomes $n$-divisible over a finite unramified extension; their compositum $L/K$ is still finite and unramified. Every class from $E(K)/nE(K)$ then maps to zero in $E(L)/nE(L)$.
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