= Solution
Expand the payoff as $(S_T-2K+K^2/S_T)1_{\{S_T>K\}}$. The same <risk-neutral measure> applies. For any real $p$, completing the square gives the <truncated lognormal moment>
$$
\mathbb E[S_T^p1_{\{S_T>K\}}]=S_0^pe^{prT+p(p-1)\sigma^2T/2}\Phi(d_2+p\sigma\sqrt T).
$$
Using $p=1,0,-1$ and discounting by $e^{-rT}$ gives
$$
\boxed{V_0=S_0\Phi(d_1)-2Ke^{-rT}\Phi(d_2)+\frac{K^2}{S_0}e^{-2rT+\sigma^2T}\Phi(d_2-\sigma\sqrt T).}
$$
The inverse-price moment explains both the extra volatility factor and the shifted normal argument in the last term.
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