= Solution
For equal prior probabilities, $b=\tfrac12a^T(\mu_1+\mu_2)$. The projected score is <normal> under either class because it is a <linear image of a multivariate normal vector>, and its <variance> is
$$
a^TVa=(\mu_1-\mu_2)^TV^{-1}(\mu_1-\mu_2)=\delta^2.
$$
Moreover $a^T\mu_1-a^T\mu_2=\delta^2$, so the midpoint threshold lies $\delta^2/2$ from either projected <mean>. Hence
$$
a^TX-b\mid C_1\sim N(\delta^2/2,\delta^2),\qquad a^TX-b\mid C_2\sim N(-\delta^2/2,\delta^2).
$$
Standardizing the two <normal distributions>, with $\Phi$ the <standard normal distribution function>, gives
$$
\boxed{P(\text{assign }C_2\mid C_1)=\Phi(-\delta/2)=P(\text{assign }C_1\mid C_2).}
$$
Thus the <equal-covariance Gaussian classification error> is $p(\delta)=\Phi(-\delta/2)$. The <Mahalanobis distance> $\delta$ between the class <means> is their separation in units of within-class noise. It yields $p\to1/2$ as $\delta\downarrow0$ and $p\to0$ as $\delta\to\infty$. The stipulated $\delta>0$ means the two class <means> are distinct.
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