Let $Q\in E(K)$. Since $E(K)/\mathfrak{p}E(K)=0$, we know that $Q\in\mathfrak{p}E(K)$. You've assumed that $\mathfrak{p}$ is principal, say $\mathfrak{p}=(\pi)$, so $Q=u_1\pi Q_1$ for some unit $u_1\in\mathcal{O}_K^*$ and some point $Q_1\in E(K)$. Repeating $n$ times, we get $Q=u_n\pi^nQ_n$. (For simplicity, I'll assume that $p=N\mathfrak{p}$ is odd.) Then the easiest way to finish the proof at this point is probably to use canonical heights to conclude that
$$ \hat h(Q)=\hat h(u_n\pi^nQ_n)=p^n\hat h(Q_n). $$
Next we observe that
$$ h(Q_n) = \hat h(Q_n) + O(1) = p^{-n}\hat h(Q) + O(1), $$
so $(Q_n)_{n\ge1}$ is a set of bounded height in $E(K)$, hence is finite. Hence we can find $n>m$ with $Q_n=Q_m$, and hence
$$ Q = u_n\pi^nQ_n = u_n\pi^nQ_m = u_nu_m^{-1}\pi^{n-m}u_m\pi^mQ_m = u_nu_m^{-1}\pi^{n-m}Q. $$ Therefore
$$ (u_nu_m^{-1}\pi^{n-m}-1)Q = 0, $$
which proves that every point in $E(K)$ is a torsion point.

(There may well be easier ways to do this, but this is the first thing that occurred to me.)