According to the answers in posting it is possible that $\mathsf{VP} = \mathsf{VNP}$ and $\mathsf{P} \neq \mathsf{NP}$ are simultaneously correct. $\mathsf{VP} = \mathsf{VNP}$ implies $\mathsf{P/poly} = \mathsf{PH/poly}$ (assuming the Generalized Riemann Hypothesis).This means that $\mathsf{VP} = \mathsf{VNP}$ would amplify the power of polynomial circuits even if $\mathsf{P} \neq \mathsf{NP}$. The best known obstruction for $\mathsf{P} \neq \mathsf{BPP}$ comes from circuit related issue for problems in $\mathsf{E}$. However if circuits have more power which is the scenario posed by $\mathsf{VP}=\mathsf{VNP}$, may be the obstruction for $\mathsf{P} \neq \mathsf{BPP}$ is not legitimate anymore.How does $\mathsf{VP} = \mathsf{VNP}$ amplify the power of randomness?If $\mathsf{VP} = \mathsf{VNP}$ were truth then is $\mathsf{P}\neq \mathsf{BPP}=\mathsf{NP}$ most likely scenario? Are other possibilities such as $$(1)\mbox{ }\mathsf{P}=\mathsf{BPP}\neq \mathsf{NP}\quad \quad(2)\mbox{ }\mathsf{P}=\mathsf{BPP}=\mathsf{NP}\quad\quad(3)\mbox{ }\mathsf{P}\neq\mathsf{BPP}\neq\mathsf{NP}$$ least likely?Are there any reasons to believe or not believe so? 
Consequences of VP = VNP on randomness
circuit complexity;counting complexity;derandomization;p vs np;hierarchy theorems