Note that if $|p|=1$, then $Y=\beta_0 + \beta_1X$. For fixed values of $X$, $Y|x = \beta_0 +\beta_1x$. Argue that $\mu_{Y|x}$ is a linear function of $x$.
Here is a question that I have recently encountered. This is my attempt in answering it.
$Y|x = \beta_0 +\beta_1x$
$\implies E(Y|x) = E(\beta_0 +\beta_1x)$ (taking the expectation of both sides)
$\implies \mu_{Y|x}=\beta_0 +\beta_1x$ (since $\beta_0 +\beta_1x$ can be treated as a constant)
and thus $\mu_{Y|x}$ is a linear function of $x$.
Is my approach flawed, or is this acceptable?
Usually, the definition of linearity w.r.t. $x$ is done via (partial) derivatives, i.e., check whether the derivative of $\mu_{Y|x}$ w.r.t. $x$ depends on $x$. Namely, \begin{align} \frac{\partial}{\partial x} E[Y|X=x] = \frac{\partial}{\partial x}( \beta_0 + \beta_1 x) = \beta_1, \end{align} which is not a function of $x$, hence your model is indeed linear.