For which values of the parameter $p$ will the set of vectors $B =\{b_1,b_2,b_3,b_4\}$ form a basis in $\mathbb R^4$? For these values of $p$ determine the coordinate vector $[v]_B$.
$$\begin{align} b_1 &= (\sqrt2,2,11,13)^T\\ b_2 &= (\sqrt2,9,15,12)^T\\ b_3 &= (\sqrt2,-19,4,19)^T\\ b_4 &=(\sqrt2,9,-5,p)^T\\ v &=(\sqrt2,-5,-48,-19+3p)^T\end{align}$$
As far as I understood we have to find these values of $p$ for which vectors $(b_1,b_2,b_3,b_4)$ will be independent and span the whole space.
On
The vectors form a basis if and only if the determinant of their matrix is non-zero.
$$\det \begin{pmatrix} \sqrt2 & \sqrt2 & \sqrt2 & \sqrt 2 \\ 2 & 9 & -19 & 9 \\ 11 & 15 & 4 & -5 \\ 13 & 12 & 19 & p \end{pmatrix} = \begin{vmatrix} \sqrt2 & 0 & 0 & 0 \\ 2 & 7 & -21 & 7 \\ 11 & 4 & -7 & -16 \\ 13 & -1 & 6 & p-13 \end{vmatrix} = \begin{vmatrix} \sqrt2 & 0 & 0 & 0 \\ 2 & 7 & 0 & 0 \\ 11 & 4 & 5 & -20 \\ 13 & -1 & 3 & p-12 \end{vmatrix} = \begin{vmatrix} \sqrt2 & 0 & 0 & 0 \\ 2 & 7 & 0 & 0 \\ 11 & 4 & 5 & 0 \\ 13 & -1 & 3 & p \end{vmatrix} = p \cdot 35\sqrt2$$
So, the answer to the first part of the question is that these vectors form a basis iff $p \ne 0$.
To answer the second part we are looking for a set of numbers $x$, $y$, $z$, $w$, such that:
$$ x \mathbf{b}_1 + y\mathbf{b}_2 + x\mathbf{b}_3 + w\mathbf{b}_4 = \mathbf{v}$$
That is, we need to solve the following system of equations:
$$\begin{align*} \sqrt2x + \sqrt2y + \sqrt2z + \sqrt2w &= \sqrt2 \\ 2x + 9y -19z + 9w &= -5 \\ 11x + 15y + 4z -5w &= -48 \\ 13x + 12y + 19z + pw &= -19 + 3p \end{align*}$$
Note that $p$ here is a parameter, not an unknown variable. Doing simple transformation (multiplying an equation by a number, adding or subtracting a multiple of an equation from another equation) we do the variable elimination:
$$\begin{align*} x + y + z + w &= 1 \\ 7y - 21z + 7w &= -7 \\ 4y - 7z -16w &= -59 \\ -y + 6z + (p-13)w &= -32 + 3p \end{align*}$$
$$\begin{align*} x + y + z + w &= 1 \\ y - 3z + w &= -1 \\ 5z -20w &= -55 \\ 3z + (p-12)w &= -33 + 3p \end{align*}$$
$$\begin{align*} x + y + z + w &= 1 \\ y - 3z + w &= -1 \\ z -4w &= -11 \\ pw &= 3p \end{align*}$$
Unraveling this from the bottom to the top we get: $w=3$, $z=1$, $y=-1$, and $x=-2$. Which means that the coordinates of $\mathbf{v}$ in this $\mathbf{B}$ basis are:
$$ [\mathbf{v}]_\mathbf{B} = (-2,-1,1,3). $$
Vectors form a basis of $\mathbb R^4$ if they are linearly independent and if they span the space. Since you have four vectors, and $\dim\mathbb R^4=4$, you just need to check their linear independence.
Vectors are linearly independent (L. I.) if the linear combination is equal to 0 only when all the coefficients are zero, that is: $$\alpha b_1+\beta b_2 +\gamma b_3+ \delta b_4=0 \iff \alpha=\beta=\gamma=\delta=0\tag1$$ $(1)$ is a system of four equations, which you can solve to find the desired value(s) of $p$.