The question is how do I find $p_{ij}^{(n)}$ given that it's true that $P^{(n)}=P^n$ where $P$ is the 1 step transition matrix, but it's not always true that $p_{ij}^{(n)}=p_{ij}^n$ (according to my notes).
It seems like that 2nd equation not holding implies it is no use to diagonalize $P$ to find $P^n$ in order to find $p_{ij}^{(n)}$.
It is true that if $P=ADA^{-1}$ for some diagonal matrix $D$ then $P^n=AD^nA^{-1}$. However, this need not be true in order to compute $P^n$. Consider
$$P = \begin{pmatrix} 5/12 & 5/12 & 1/6 \\ 1/4 & 1/4 & 1/2 \\ 1/3 & 1/3 & 1/3 \end{pmatrix},$$ then $$ P^2 = \left( \begin{array}{ccc} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \end{array} \right) $$ and hence $$P^n=\left( \begin{array}{ccc} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \end{array} \right) $$ for any $n\geqslant 2$.