 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 $\tilde{E}(\overline{k})$. Choose a point $\tilde{Q}$ with $[n]\tilde{Q}=\tilde{P}$ 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 formSince $n$ 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 $P_1,\dots ,P_s$. 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 $E(L)/nE(L)$.