It is possible to project without an orthonormal basis.
Our space of interest $W$ is the column space of the matrix $A$ given by
$$
A=\left[\begin{array}{rrrr}
1 & 1 & 0 & 1 \\
-3 & 5 & 4 & -2 \\
0 & 2 & 1 & 0 \\
1 & 3 & 1 & 4
\end{array}\right]
$$
A basis of $W=\operatorname{Col}(A)$ is given by the columns of $A$ corresponding to pivot columns in $\operatorname{rref}(A)$. The reduced row echelon form of $A$ is
$$
\operatorname{rref}(A)=\left[\begin{array}{rrrr}
1 & 0 & -\frac{1}{2} & 0 \\
0 & 1 & \frac{1}{2} & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
This shows that $A$ has rank three and that the vectors
\begin{align*}
\left\langle1,\,-3,\,0,\,1\right\rangle &&
\left\langle1,\,5,\,2,\,3\right\rangle &&
\left\langle1,\,-2,\,0,\,4\right\rangle
\end{align*}
form a basis of $W$.
Now, put these basis vectors into the columns of a matrix
$$
X = \left[\begin{array}{rrr}
1 & 1 & 1 \\
-3 & 5 & -2 \\
0 & 2 & 0 \\
1 & 3 & 4
\end{array}\right]
$$
The projection matrix onto $W$ is
$$
P=X(X^\top X)^{-1}X^\top
$$
In our case, we end up with
$$
P=\left[\begin{array}{rrrr}
\frac{131}{231} & -\frac{10}{77} & \frac{10}{21} & \frac{10}{231} \\
-\frac{10}{77} & \frac{74}{77} & \frac{1}{7} & \frac{1}{77} \\
\frac{10}{21} & \frac{1}{7} & \frac{10}{21} & -\frac{1}{21} \\
\frac{10}{231} & \frac{1}{77} & -\frac{1}{21} & \frac{230}{231}
\end{array}\right]
$$
We can now project any vector $v\in\Bbb R^4$ onto $W$ by computing $Pv$.
If you're dead set on using orthonormal bases, then you could take these three basis vectors and apply the Gram-Schmidt algorithm. This gives the new basis
\begin{align*}
q_1 &=
\left\langle\frac{1}{\sqrt{11}},\,-\frac{3}{\sqrt{11}},\,0,\,\frac{1}{\sqrt{11}}\right\rangle &
q_2 &=
\left\langle\frac{1}{\sqrt{7}},\,\frac{1}{\sqrt{7}},\,\frac{1}{\sqrt{7}},\,\frac{2}{\sqrt{7}}\right\rangle &
q_3 &=
\left\langle-\frac{1}{\sqrt{3}},\,0,\,-\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right\rangle
\end{align*}
We can then project any $v\in\Bbb R^4$ with the formula
$$
Pv=\langle q_1, v\rangle q_1 + \langle q_2, v\rangle q_2 + \langle q_3, v\rangle q_3
$$
Best Answer
Pick any point $x_{0}$ on the hyperplane $H=\{ x : \langle x, a \rangle = b \}$. Then $$ H-x_{0} = \{ x-x_{0} : x \in H \} $$ is a subspace because $\langle (x-x_{0}),a \rangle = 0$. So your hyperplane is a translation of a subspace in a particular vector direction. If you want to project $y$ onto $H$, that's the same as projecting $y-x_{0}$ onto the subspace $H-x_{0}$, and then adding $x_{0}$ back because the final answer to this modified projection problem is on $H-x_{0}$. Translation by a fixed vector is an isometric operation (i.e., it keeps relative distances the same.) That guarantees that the one projection problem is equivalent to the other. In other words $\|x-y\|=\|(x-x_{0})-(y-x_{0})\|$.
So, suppose you have found such an $x_{0}$ for which $\langle x_{0}, a\rangle = b$, and you now want to project $y$ onto $H$. Projecting $y-x_{0}$ onto $H-x_{0}$ gives $$ (y-x_{0}) - \frac{\langle y-x_{0},a\rangle}{\|a\|^{2}}a =y-x_{0}+\frac{\langle y,a\rangle - b}{\|a\|^{2}}a $$ The $b$ appeared because $x_{0}$ was chosen so that $\langle x_{0},a\rangle =b$. Then to put everything back into the original coordinates, you add $x_{0}$ back to your final answer because the above is a point on $H-x_{0}$. So the original projection, in the original coordinates, is found to be $$ y - \frac{\langle y,a\rangle-b}{\|a\|^{2}}a. $$