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Every finite field extension is a Galois extension, and

Let $K^{\mathrm{ur}}$ be the maximal unramified extension of $K$, with residue field $\overline{k}$. First show that every geometric solution of $nQ=P$ lies in $E(K^{\mathrm{ur}})$.

The multiplication-by-n morphism on the reduced elliptic curve is surjective over $\overline{k}$. Choose $\overline{Q}$ with $n\overline{Q}=\overline{P}$, defined over some finite finite field extension $k^{\prime }/k$. Let $L/K$ be the corresponding finite unramified extension. By the surjectivity of good reduction over a local field, lift $\overline{Q}$ to $Q_0\in E(L)$. Then $P-n{Q}_0\in E_1(L)$. The multiplication isomorphism of a formal group law gives a unique $R\in E_1(L)$ with $nR=P-n{Q}_0$, so $Q_0+R$ is an $n$-division point of $P$ defined over $L$.

Taking $P=O$ gives every element of $E[n]$ over $K^{\mathrm{ur}}$: reduction is injective on this torsion point of an elliptic curve group because $[n]$ is injective on the formal kernel, and each reduced $n$-torsion point has the unique corrected lift just constructed. Every other division point differs from one lift by an element of $E[n]$. Hence the entire field $K([n{]}^{-1}P)$ is unramified, not merely a field containing one choice of $Q$.

To bound the composite uniformly in $P$, put $f=[K(E[n]):K]$. This division field of an elliptic curve is unramified. The Galois group of $K^{\mathrm{ur}}/K$ is procyclic, generated topologically by its Frobenius element $\sigma$, and ${\sigma }^f$ fixes every element of $E[n]$. If $nQ=P\in E(K)$, thenIteration gives ${\sigma }^{jf}Q=Q+jT$. In particular ${\sigma }^{nf}Q=Q$. Thus every such $Q$, for every $P$ together, lies in the unique unramified extension of degree $nf$. If $M$ is the composite of all these fields, the uniform unramified division field over a local field givesThe prime-to-$p$ hypothesis is essential to the inverse-series argument; no assertion about arbitrary $p$-division is being made.