Note: your previous question assumed log-normality instead of normality.
By Cholesky decomposition, we assume that, under measure $P$,
\begin{align*}
\frac{dR_t}{R_t} &= \mu_{R,t} dt + \sigma_{R,t}\, dW^1_t\\
\frac{dA_t}{A_t} &= \mu_{A,t} dt + \sigma_{A,t}\, d\left(\rho W^1_t + \sqrt{1-\rho^2} W^2_t\right),
\end{align*}
where $W^1$ and $W^2$ are two independent standard Brownian motions. Here, we assume that $\mu_{R,t}$, $\mu_{A,t}$, $\sigma_{R,t}$, and $\sigma_{A,t}$ are deterministic or constants.
Define the probability measure $\widetilde{P}$ such that we have the Radon Nykodym derivative
\begin{align*}
\frac{d\widetilde{P}}{dP}\big|_t &= \frac{A_t}{A_0}\frac{1}{e^{\int_0^t\mu_ {A,s}ds}}\\
&= \exp\left(-\frac{1}{2}\int_0^t\sigma_{A,s}^2 ds + \int_0^t\sigma_{A,s}\,d\left(\rho W^1_s + \sqrt{1-\rho^2} W^2_s\right)\right).
\end{align*}
By Girsanov theorem,
\begin{align*}
\widetilde{W}^1_t &= W_t^1 - \rho \int_0^t \sigma_{A,s} ds \,\, \mbox{ and}\\
\widetilde{W}^2_t &= W_t^2 - \sqrt{1-\rho^2} \int_0^t \sigma_{A,s} ds
\end{align*}
are two independent standard Brownian motions under $\widetilde{P}$. Moreover, under $\widetilde{P}$,
\begin{align*}
\frac{dR_t}{R_t} &= \left(\mu_{R,t}+\rho \sigma_{A,t} \sigma_{R,t}\right) dt + \sigma_{R,t}\, d\widetilde{W}^1_t.
\end{align*}
Note also that
\begin{align*}
\frac{dP}{d\widetilde{P}}\big|_t &= \frac{A_0}{A_t}e^{\int_0^t\mu_ {A,s}ds}.
\end{align*}
Then,
\begin{align*}
E_P(A_T(R_T - \lambda)^+) &= E_{\widetilde{P}}\left(\frac{dP}{d\widetilde{P}}\big|_T A_T (R_T - \lambda)^+ \right)\\
&= A_0\, e^{\int_0^T\mu_{A, s} ds }\,E_{\widetilde{P}}\left( (R_T - \lambda)^+ \right),
\end{align*}
which reduces to Black-Scholes' formula. Some details are omitted here.
If $\mu_{A,t}$ is the short interest rate at time $t$, then $e^{\int_0^t\mu_ {A,s}ds}$ is the money-market account value at time $t$.
