Let $\theta\sim\mathcal{N}(\bar{\theta},1/\tau_\theta)$ be a normally distributed random variable, $\varepsilon\sim\mathcal{N}(0,1/\tau_\varepsilon)$ be a normally distributed noise term independent from $\theta$, and let $P$ be an affine function of $\theta$ and $\varepsilon$, i.e., $P(\theta,\varepsilon)=a+b\theta+c\varepsilon$ with $b,c\neq 0$.
I want to determine the conditional variance $Var(\theta+P|P=p)$.
I know how to determine the conditional variance $Var(\theta|P=p)$, which should be $Var(\theta|P)=b^2/\tau_\theta+c^2/\tau_\varepsilon$, since $\theta$ and $\varepsilon$ are uncorrelated. I am not entirely sure about $Var(\theta+P|P=p)$ though, since $P$ and $\theta$ are not uncorrelated.
Here is a guess: \begin{equation} Var(\theta+P|P=p)=Var(\theta|P=p)+Var(P|P=p)+2Cov(\theta,P|P=p)\\ =Var(\theta|P=p)+2Cov(\theta,P|P=p) \end{equation} Now I am not sure; since conditional on $P=p$, $P$ does not vary, the covariance term $$2Cov(\theta,P|P=p)=2E[(\theta-E[\theta|P=p])(P-E[P|P=p])|P=p]$$ should be zero, which implies $Var(\theta+P|P=p)=Var(\theta|P=p)$.
Another approach that seems plausible would be \begin{equation} Var(\theta+P|P=p)=Var(\theta+a+b\theta+c\varepsilon|P=p)\\ =Var(\theta(1+b)+c\varepsilon|P=p)\\ \neq Var(\theta|P=p) \end{equation}
Which is it, or is it something else entirely? If it is the latter, what is $Var(\theta(1+b)+c\varepsilon|P=p)$ explicitly? Thanks!