derivatives - How to understand the market price of risk

Consider the stochastic vol:

$$dS = \mu Sdt + \sigma SdW_1$$ $$d\sigma = p(\sigma,S,t)dt + q(\sigma,S,t)dW_2$$ $$dW_1dW_2 = \rho dt$$ We want to obtain the price of option $V(\sigma,S,t),$ we use the underlying asset $S$ and another option $V_1(\sigma,S,t)$ to build the hedging portfolio: $$\Pi = V -\Delta S - \Delta_1 V_1$$ then make $$d \Pi = r\Pi dt$$ eliminate the risk terms we have $$\dfrac{\partial V}{\partial t} + \dfrac{1}{2}\sigma^2S^2\dfrac{\partial^2 V}{\partial S^2} + \rho\sigma Sq\dfrac{\partial^2 V}{\partial S\partial \sigma} + \dfrac{1}{2}\sigma^2q^2\dfrac{\partial^2 V}{\partial \sigma^2} + rS\dfrac{\partial V}{\partial S} -rV = -(p-\lambda q)\dfrac{\partial V}{\partial \sigma}.$$ Here $\lambda$ is called market price of risk, since we can understand $\lambda$ as $$d V -rV dt = q\dfrac{\partial V}{\partial S}(\lambda d t + d W_2) = q\Delta(\lambda d t + d W_2)$$ this is the unit of extra return.

And we have another way to price $V,$ the discounted value of $V$ is martingale, namely $dt$ term of $d(e^{-rt} V)$ is zero, then we find that, the PDE of $V$ is exactly

$$\lambda = 0$$ in above PDE. So does that mean, the discounted value of $V$ is martingale is equivalent to the market price of risk is zero?