Real jordan form to complex jordan form then compute P matrix. I have the matrix 
$$A = 
\begin{bmatrix}
5 &  0 & 1 &  0 & 0 & -6 \\
3 & -1 & 3 &  1 & 0 & -6 \\
6 & -6 & 5 &  0 & 1 & -6 \\
7 & -7 & 4 & -2 & 4 & -7 \\
6 & -6 & 6 & -6 & 5 & -6 \\
2 &  1 & 0 &  0 & 0 &  0
\end{bmatrix}$$
This can be brought in the following Jordan form, i.e. $A = TJT^{-1}$.
$$J = 
\begin{bmatrix}
2-3j & 1 & 0 & 0 & 0 & 0 \\
0 & 2-3j & 1 & 0 & 0 & 0 \\
0 & 0 & 2-3j & 0 & 0 & 0 \\
0 & 0 & 0 & 2+3j & 1 & 0 \\
0 & 0 & 0 & 0 & 2+3j & 1 \\
0 & 0 & 0 & 0 & 0 & 2+3j 
\end{bmatrix}$$
$$T = 
\begin{bmatrix}
2j & 2j & 1+j & -2j & -2j & 1-j \\
1+j & 2j & j & 1-j & -2j & -j \\
0 & 2j & 2j & 0 & -2j & -2j \\
0 & 1+j & 2j & 0 & 1-j & -2j \\
0 & 0 & 2j & 0 & 0 & -2j \\
-1+j & -1+j & j & -1-j & -1-j & -j
\end{bmatrix} 
$$
Now I have to bring A into its real Jordan form. This is easy:
$$J^{R} = 
\begin{bmatrix}
2 & 3 & 1 & 0 & 0 & 0 \\
-3 & 2 & 0 & 1 & 0 & 0 \\
0 & 0 & 2 & 3 & 1 & 0 \\
0 & 0 & -3 & 2 & 0 & 1 \\
0 & 0 & 0 & 0 & 2 & 3 \\
0 & 0 & 0 & 0 & -3 & 2  
\end{bmatrix}
$$
Now I have to compute $V$ such that $VJ^RV^{-1} = A$. My question now is how do I compute this $V$? For real valued jordan forms this is easy. I just have to compute the eigenvectors of A, $\{T_i\}$ and then $T = \begin{bmatrix} T_1 | T_2 | \ldots | T_n\end{bmatrix}$.

Notice that, with every complex pair of eigenvalues $\lambda = a \pm ib$, there exists a complex pair of eigenvectors $u \pm i v$. If you look at the columns of your matrix $T$, you can observe that you can pair up your eigenvectors according to complex conjugates in this precise way.
In real canonical form, each of your real Jordan blocks $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ comes directly from the eigenspaces generated by the eigenvectors $u \pm iv$. So your $V$ should look like
$$V = [v_1 | u_1 | v_2 | u_2 | v_3 | u_3]$$
