= Solution
\b[Only the prime two can ramify.] More precisely, every field $\mathbb Q(P_n)$ is <unramified> at every odd <prime>; actual ramification at two depends on $P$ and the chosen division point. Use $[2^n]P_n=P$, as printed in the original PDF.
Fix an odd prime $p$ and put $N=2^n$. The <elliptic-curve discriminant> $-2^8$ gives <good reduction> at $p$, and $N$ is a unit in $\mathbb Z_p$. The <unramified division torsors at good primes> imply that all <division points of an elliptic curve> above $P$ are defined over finite <unramified extensions> of $\mathbb Q_p$. Here is a direct lifting proof, which does not require $P$ to have integral affine coordinates.
Properness gives a reduction $\widetilde P$. Choose a point $\overline Q$ with $[N]\overline Q=\widetilde P$ over the algebraic closure of the <residue field>, and a finite residue extension containing it. Pass to the corresponding finite <unramified extension> $K/\mathbb Q_p$. The <surjectivity of good reduction over a local field> lifts $\overline Q$ to $Q_0\in E(K)$. Its error $[N]Q_0-P$ lies in the <kernel of reduction of an elliptic curve>. The <prime-to-residue-characteristic multiplication on a formal group> makes $[N]$ bijective on that kernel. Choose the unique kernel point $R$ with $[N]R=[N]Q_0-P$; then $Q=Q_0-R$ satisfies $[N]Q=P$ and has reduction $\overline Q$.
Every algebraic division point has some such reduction and is the unique lift in its division fibre: the difference of two lifts lies in the reduction kernel, which has no $N$-torsion. Thus every chosen $P_n$ lies in an unramified local extension. For every prime of its possibly non-Galois coordinate field above $p$, the <ramification index> is one. Globally,
$$
\boxed{\mathbb Q(P_n)/\mathbb Q\text{ is unramified outside }\{2\}.}
$$
One cannot universally assert that two actually ramifies: $P=O$ and $P_n=O$ give the field $\mathbb Q$. It can ramify. For example, a half of $T=(-1,0)$ is
$$
Q=\left(-1+\sqrt2,\ 2\sqrt{\sqrt2-1}\right),\qquad [2]Q=T.
$$
The <elliptic curve group law> verifies this by doubling. Its coordinate field contains $\mathbb Q(\sqrt2)$, which has <ramification in the quadratic extension generated by the square root of two>, so the only possible bad prime does occur for this choice.
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