I check it with the standard Garlerkin method and confirmed that it is right, in both cases (i) and (ii).
Discribe the proof briefly (under (ii)):
Take a CONS of $H$ as $h_1 = 1$ and $h_k(x) = \cos[(k-1)\pi x]$.
First notice that if $V(u) = V^\dagger(\langle u, h_1 \rangle, \ldots, \langle u, h_N \rangle)$ and $\langle B(u), h_k \rangle = B_k^\dagger(\langle u, h_1 \rangle, \ldots, \langle u, h_N \rangle)$ for some $V^\dagger$, $B_k^\dagger \in C_b^2(\mathbb R^N)$ for $k \leq N$, and $B_k^\dagger = 0$ for $k > N$, then it becomes an $N$-dimensional diffusion and a infinite-dimensional OU process, and the result is obvious.
Then take $V_N^\dagger$ and $B_{N, k}^\dagger$ be the marginal expectation, i.e., for $x \in \mathbb R^N$ and $k \leq N$, $$V_N^\dagger(x) = \int_{\mathbb R^\infty} V\left(\sum_{1}^N x_kh_k + \sum_{N+1}^\infty y_kh_k\right)\prod_{N+1}^\infty \Phi_k(y_k)dy_k, $$ $$B_{N, k}^\dagger(x) = e^{2V_N^\dagger(x)} \int_{\mathbb R^\infty} U_k\left(\sum_{1}^N x_kh_k + \sum_{N+1}^\infty y_kh_k\right)\prod_{N+1}^\infty \Phi_k(y_k)dy_k, $$ where $U_k = e^{-2V}\langle B, h_k \rangle$ and $\Phi_k(x) = \sqrt{\lambda_k\pi^{-1}}e^{-\lambda_kx^2}$. This finite-dimensional approximation protect the condition (ii), so we have the invariant measure, due to the previous step. Finally take the limit.