= Solution
Let $K^{\rm ur}$ be the <maximal unramified extension> of $K$, with <residue field> $\overline k$. First show that every geometric solution of $nQ=P$ lies in $E(K^{\rm ur})$.
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> $k'/k$. 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 $P-nQ_0\in E_1(L)$. The <multiplication isomorphism of a formal group law> gives a unique $R\in E_1(L)$ with $nR=P-nQ_0$, so $Q_0+R$ is an $n$-division point of $P$ defined over $L$.
Taking $P=O$ gives every element of $E[n]$ over $K^{\rm ur}$: 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 $K^{\rm ur}/K$ is procyclic, generated topologically by its Frobenius element $\sigma$, and $\sigma^f$ fixes every element of $E[n]$. If $nQ=P\in E(K)$, then
$$
T=\sigma^fQ-Q\in E[n],\qquad \sigma^fT=T.
$$
Iteration 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> gives
$$
\boxed{M/K\text{ is finite unramified},\qquad [M:K]\leq n[K(E[n]):K].}
$$
The prime-to-$p$ hypothesis is essential to the inverse-series argument; no assertion about arbitrary $p$-division is being made.
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