The$\newcommand\la\lambda\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand\R{\mathbb R}\newcommand{\Si}{\Sigma}$The answer is negative, even for $k=3$.
A counterexample is as follows: $$\lambda=2,\quad u=\frac{(4,-3,-3)}{\sqrt{34}},$$ $$e_1:=(1,1,1)/\sqrt3,\quad e_2:=(-1,1,1)/\sqrt3.$$ Then $e_1\cdot u>e_2\cdot u$, whereas $p_1=0.100\ldots<0.183\ldots=p_2$.
The OP has changed the question, replacing the comparisons of the values of $e_i\cdot u$ by the comparisons of the values of $(e_i\cdot u)^2$.
Then the answer becomes positive, at least for $k=3$.
Indeed, if $X_1,X_2,X_3$ are jointly Gaussian zero-mean random variables with correlation coefficients $\rho_{ij}=Corr(X_i,X_j)$, then, by the 3rd display from the bottom of p. 355 of [Plackett's paper], \begin{equation*} P(X_1>0,X_2>0,X_3>0) \\ =\frac1{4\pi}\,(2\pi-\cos^{-1}\rho_{12}-\cos^{-1}\rho_{23}-\cos^{-1}\rho_{31}). \end{equation*}
If $a=(a_1,a_2,a_3)$ is an eigenvector of the covariance matrix $\Si$ of the random vector $X=(X_1,X_2,X_3)$ corresponding to the eigenvalue $t:=\la>1$, with the other two eigenvalues each equal $1$, then \begin{equation*} \Si=\frac1{a_1^2+a_2^2+a_3^2} \left( \begin{array}{ccc} a_1^2 t+a_2^2+a_3^2 & a_1 a_2 (t-1) & a_1 a_3 (t-1) \\ a_1 a_2 (t-1) & a_2^2 t+a_1^2+a_3^2 & a_2 a_3 (t-1) \\ a_1 a_3 (t-1) & a_2 a_3 (t-1) & a_3^2 t+a_1^2+a_2^2 \\ \end{array} \right). \end{equation*}
It follows that for any $\ep=(\ep_1,\ep_2,\ep_3)\in\{-1,1\}^3$ \begin{equation*} \begin{aligned} p_{a,\ep}:=4\pi P(\ep_1 X_1>0,\ep_2 X_2>0,\ep_3 X_3>0)-2\pi \\ =-\cos ^{-1}R_{12}-\cos ^{-1}R_{23}-\cos ^{-1}R_{31}, \end{aligned} \end{equation*} where \begin{equation*} R_{ij}:=\frac{(t-1)a_i a_j \ep_i \ep_j } {\sqrt{\big((t-1)a_i^2+a_1^2+a_2^2+a_3^2\big) \big((t-1)a_j^2+a_1^2+a_2^2+a_3^2\big) }}. \end{equation*}
We want to show that for any $a=(a_1,a_2,a_3)\in\R^2\setminus\{0\}$ and any $\de=(\de_1,\de_2,\de_3)$ and $\ep=(\ep_1,\ep_2,\ep_3)$ in $\{-1,1\}^3$ such that \begin{equation*} (a\cdot\de)^2\le(a\cdot\ep)^2 \tag{1}\label{1} \end{equation*} we have \begin{equation*} p_{a,\de}\le p_{a,\ep}. \tag{2}\label{2} \end{equation*}
Note that $p_{a,\ep}$ is even in $a$ and in $\ep$ ($p_{-a,\ep}=p_{a,\ep}=p_{a,-\ep}$), and $(a\cdot\ep)^2$ is even in $a$ and in $\ep$ as well. Also, $p_{a,\ep}$ is permutation invariant: $p_{a,\ep}$ remains invariant if the indices of $a$ and $\ep$ are permuted by the same permutation. So, without loss of generality (wlog) \begin{equation*} \de=(-1,1,1)\quad\text{and}\quad\ep=(1,1,1), \end{equation*} and then conditions \eqref{1} and \eqref{2} can be rewritten as \begin{equation*} a_1(a_2+a_3)\ge0\tag{1a}\label{1a} \end{equation*} and \begin{equation*} \sin^{-1}S_{12}+\sin^{-1}S_{13}\ge0, \tag{2a}\label{2a} \end{equation*} where \begin{equation*} S_{ij}:=S_{ij}(a):=\frac{(t-1)a_i a_j } {\sqrt{\big((t-1)a_i^2+a_1^2+a_2^2+a_3^2\big) \big((t-1)a_j^2+a_1^2+a_2^2+a_3^2\big) }}. \end{equation*} Also, $a_1(a_2+a_3)$ and the $R_{ij}$'s are even in $a$, and the $S_{ij}$'s are homogeneous in $a$: $S_{ij}(ka)=S_{ij}(a)$ for any real $k\ne0$. Also, by continuity, wlog the inequality in \eqref{1a} is strict. So, wlog $a_1=1$ and $a_2+a_3>0$.
So, wlog $a_2=a$ and $a_3=b$ for some real $a$ and $b$ such that $a>0$ and $a+b>0$.
If $b\ge0$, then $S_{12}\ge0$ and $S_{13}\ge0$, so that \eqref{2a} is obvious.
If $b<0$, then $-a<b<0$ and hence $b^2<a^2$. Then ($S_{12}>0$ and) \begin{equation*} r(t):=\frac{S_{13}^2}{S_{12}^2}=\frac{b^2 \left(a^2 t+b^2+1\right)}{a^2 \left(a^2+b^2 t+1\right)}, \end{equation*} which is increasing from $r(1+)=\frac{b^2}{a^2}<1$ to $r(\infty-)=1$ as $t$ is increasing from $1$ to $\infty$, so that $|S_{13}|<S_{12}$ and \eqref{2a} follows. $\quad\Box$