Answer
$$
P^{-1}= \left[ \begin {array}{cccccccc} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1
\\ 0&0&0&1\end {array} \right].
$$
Work Step by Step
Given
$$
B=\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}, \\ B^{\prime}= \{(1,1,1,1),(0,1,1,1),(0,0,1,1),(0,0,0,1)\} .
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccccc} 1&1&1&1&1&0&0&0\\ 0
&1&1&1&0&1&0&0\\ 0&0&1&1&0&0&1&0
\\ 0&0&0&1&0&0&0&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{4}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccccc} 1&0&0&0&1&-1&0&0\\ 0
&1&0&0&0&1&-1&0\\ 0&0&1&0&0&0&1&-1
\\ 0&0&0&1&0&0&0&1\end {array} \right]
.$$
So, we have
$$
P^{-1}= \left[ \begin {array}{cccccccc} 1&-1&0&0\\ 0&1&-1&0\\ 0&0&1&-1
\\ 0&0&0&1\end {array} \right].
$$