Assuming you mean $P_t=S_T$, that you are pricing under the risk neutral measure $\mathbb{Q}$ and introducing a discount factor $e^{-\int_0^T{r(t)dt}}$, your equation can be rewritten $-$ where $\{\mathcal{F}_n\}_{n\geq0}$ is an appropriate filtration:
$$ \begin{align}
\mathbb{E}^{\mathbb{Q}}\left[V_T|\mathcal{F}_0\right]=\mathbb{E}_0^{\mathbb{Q}}\left[V_T\right]&=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\max(-K,S_T-K)\right]
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&=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\left(\max(0,S_T)-K\right)\right]
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&=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\max(0,S_T)\right]-KP(0,T)
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&=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}S_T\right]-KP(0,T)
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&=S_0-KP(0,T)
\end{align} $$
Where $P(0,T)$ is the price of a zero-coupon bond of maturity $T$ $-$ by definition equal to:
$$ P(0,T) = \mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\right] $$
If now you have two distinct values $K_1 \not= K_2$:
$$ \mathbb{E}_0^{\mathbb{Q}}\left[V_T\right]=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\max(-K_1,S_T-K_2)\right] $$
Using equivalent arguments as above, the value can be rewritten:
$$ \mathbb{E}_0^{\mathbb{Q}}\left[V_T\right]=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\max(0,S_T-(K_2-K_1))\right]-K_1P(0,T) $$
This 2nd risk neutral expectation is simply the price of a European call option of strike $K_2-K_1$ so you can use standard formulas to price it.
If you now introduce time-dependency to your "strike" $K$ $-$ as one of your comments below does $-$, we get:
$$ \begin{align}
\mathbb{E}_0^{\mathbb{Q}}\left[V_T\right]&=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\max(-K_T,S_T-K_T)\right]
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&=\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}\max(0,S_T)\right]-\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}K_T\right]
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&=S_0-\mathbb{E}_0^{\mathbb{Q}}\left[e^{-\int_0^T{r(t)dt}}K_T\right]
\end{align} $$
Here the price and the complexity of the calculations will depend on the dynamics of $K_t$, but in a Black-Scholes world, assuming a geometric Brownian motion diffusion process for $K_t$ and that this asset is tradeable, we simply get:
$$ \mathbb{E}_0^{\mathbb{Q}}\left[V_T\right]=S_0-K_0 $$
Because the discounted price of any tradeable asset under the risk neutral measure is a martingale.
The takeaway is that the complexity of the solution and the numerical techniques needed if no closed-form exists depend on the setting you specify $-$ i.e. diffusion equation followed by your asset, specification of the interest-rate process, etc. In a Black-Scholes framework, the price of the derivatives above are readily derived.