Answer
$$
P^{-1}=\left[ \begin {array}{cccccccc} -24&-10&-29&12
\\ 7&3&7&-3\\ 1&0&
3&-1\\ -2&-1&-2&1\end {array} \right].
$$
Work Step by Step
Given
$$
B=\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}, \\ B^{\prime}= \{(1,3,2,-1),(-2,-5,-5,4),(-1,-2,-2,4),(-2,-3,-5,11) \} .
$$
To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix
$$
\left[B^{\prime} B\right]= \left[ \begin {array}{cccccccc} 1&3&2&-1&1&0&0&0\\
-2&-5&-5&4&0&1&0&0\\ -1&-2&-2&4&0&0&1&0
\\ -2&-3&-5&11&0&0&0&1\end {array} \right]
.$$
Using Gauss-Jordan elimination to obtain the transition matrix
$$
\left[\begin{array}{ll}{I_{4}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccccc} 1&0&0&0&-24&-10&-29&12
\\ 0&1&0&0&7&3&7&-3\\ 0&0&1&0&1&0&
3&-1\\ 0&0&0&1&-2&-1&-2&1\end {array} \right]
.$$
So, we have
$$
P^{-1}=\left[ \begin {array}{cccccccc} -24&-10&-29&12
\\ 7&3&7&-3\\ 1&0&
3&-1\\ -2&-1&-2&1\end {array} \right].
$$
