An Example of that the Projection Operator is Independent of the Chosen Basis of the Subspace

$\newcommand{\tr}[1]{\text{Tr}\left\{#1\right\}}$ $\newcommand{\ket}[1]{\left|#1\right\rangle}$ $\newcommand{\bra}[1]{\left\langle#1\right|}$ $\newcommand{\braket}[2]{\left\langle#1\right| \left.#2\right\rangle}$ $\newcommand{\sandwich}[3]{\left\langle#1\right|#2 \left|#3\right\rangle}$ $\newcommand{\span}[1]{\text{Span}\left\{#1\right\}}$ $\newcommand{\proj}[2]{\text{Proj}_{#1}\left(#2\right )}$

Say we want to find the projection operator onto the $\mathcal{V}$ $= xy$-plane. If one uses the simplest orthonormal basis as $B = \left\{ \vec{x}, \vec{y} \right\}$. $\vec{x} = \begin{pmatrix}1\\0\\0\end{pmatrix}$, $\vec{y} = \begin{pmatrix}0\\1\\0\end{pmatrix}$, the projection operator becomes $\proj{\mathcal{V}}{\vec{v}}$ $\equiv M_\mathcal{V}\vec{v}$.   $$\begin{align}  M_\mathcal{V} &= \ket{x}\bra{x} + \ket{y}\bra{y} \\ &= \vec{x}\vec{x}+\vec{y}\vec{y} \\ &= \begin{pmatrix}1\\0\\0\end{pmatrix}\begin{pmatrix}1&0&0\end{pmatrix}+ \begin{pmatrix}0\\1\\0\end{pmatrix}\begin{pmatrix}0&1&0\end{pmatrix}\\ &= \begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} +\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix} \\ &= \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix} \end{align}$$
Lets pick another basis $B^\prime = \left\{ \vec{u}_1, \vec{u}_2 \right\}$. $\vec{u}_1 = \begin{pmatrix}\cos{\theta}\\ \sin{\theta}\\0\end{pmatrix}$, $\vec{u}_2 = \begin{pmatrix}-\sin{\theta}\\ \cos{\theta}\\0\end{pmatrix}$.  $$\begin{align}  M_\mathcal{V} &= \ket{u_1}\bra{ u_1 } + \ket{ u_2}\bra{ u_2} \\ &= \vec{u}_1\vec{u}_1+\vec{u}_2\vec{u}_2 \\ &= \begin{pmatrix}\cos{\theta}\\ \sin{\theta}\\0\end{pmatrix} \begin{pmatrix}\cos{\theta}& \sin{\theta}&0\end{pmatrix} + \begin{pmatrix}-\sin{\theta}\\ \cos{\theta}\\0\end{pmatrix} \begin{pmatrix}-\sin{\theta}& \cos{\theta}&0\end{pmatrix}\\ &= \begin{pmatrix} \cos{\theta}^2 & \cos{\theta}\sin{\theta} &0\\ \cos{\theta}\sin{\theta} & \cos{\theta}^2 &0\\0&0&0\end{pmatrix} +\begin{pmatrix} \sin{\theta}^2 & -\cos{\theta}\sin{\theta} &0\\ -\cos{\theta}\sin{\theta} & \sin{\theta}^2 &0\\0&0&0\end{pmatrix} \\ &= \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix} \end{align}$$
As we see, the answer is independent of $\theta$ hence we get the same projection operator for all possible orthonormal basis that span the $xy$-plane.

In QM, to get the probabilities we calculate the square of the norm of a vector, $\| \vec{v}_\mathcal{V} \|^2$. It can either be found by first finding the projection operator, then getting the projected component and then getting its norm etc. The end result is sandwiching the projection operator with the state (expectation value of the projection) $\sandwich{\psi}{M_\mathcal{V}}{\psi}$. Or by finding the projected vector's components on the orthonormal basis first by taking the inner product of the vector with each element of the basis.

Say $\vec{r}_\mathcal{V}$ is the projection of a vector $\vec{r}$ onto $xy$-plane. Let us express it in two different basis (corresponding to two different $\theta$ values) that span onto the $xy$-plane. $\vec{r}_{\mathcal{V},u_1}$ $= \braket{u_1}{r}$ $=\vec{u}_1\cdot \vec{r}$. $\vec{r}_{\mathcal{V},u_2}$ $= \braket{u_2}{r}$ $=\vec{u}_2\cdot \vec{r}$. And same relations for $\vec{w}_i$s. The norm square is $$\begin{align} \| \vec{r}_\mathcal{V} \|^2 & = \braket{r_\mathcal{V}}{r_\mathcal{V}} \\ &= \| \vec{r}_{\mathcal{V},u_1} \|^2 + \| \vec{r}_{\mathcal{V},u_2} \|^2 \\ &= \left| \braket{u_1}{r} \right|^2 + \left| \braket{u_2}{r} \right|^2 \\ &= \braket{r}{u_1}\braket{u_1}{r} + \braket{r}{u_2}\braket{u_2}{r} \\ &= \bra{r} \left( \ket{u_1}\bra{u_1} + \ket{u_2}\bra{u_2} \right) \ket{r} \\ &\equiv \sandwich{r}{M_\mathcal{V}}{r} \end{align}$$
Pythagorean law applies here. For any basis the square of the sides will be equal to the square of the hypotenuse, which is the square norm that we are looking. Therefore these two methods of probability calculation are equivalent. $\left| \braket{u_1}{\psi} \right|^2 + \left| \braket{u_2}{\psi} \right|^2$ $\equiv \sandwich{\psi}{M_\mathcal{V}}{\psi}$. According to Euclidean geometry this is correct for any dimensional subspaces. $\sum_i^K \left| \braket{u_i}{\psi} \right|^2 = \sandwich{\psi}{M_\mathcal{V}}{\psi}$.