By definition the symmetric part of the $(0, p)$ rank tensor $\mathcal T^{0}_p$
$$S_{p}T=\frac{1}{p!}\underset{\sigma}{\sum} P_{\sigma}T$$
$P_{\sigma}$ is $\sigma$-transpose operator. My question is how to understand this formula intuitively? $T$ is a given tensor. Is this about to construct a symmetric tensor from the elements of a given tensor? How to exploit the formula for the given $(0, 2)$ rank tensor below(the bottom one gives the numerical values of the elements of the top one)? The square matrices are considered.
\begin{bmatrix} T_{11} & T_{12} & T_{13} & T_{14}\\ T_{21} & T_{22} & T_{23} & T_{24}\\ T_{31} & T_{32} & T_{33} & T_{34}\\ T_{41} & T_{42} & T_{43} & T_{44}\\ \end{bmatrix}
\begin{bmatrix} 1 & 2 & 3 & 4\\ 8 & 7 & 6 & 5\\ 4 & 3 & 2 & 1\\ 5 & 6 & 7 & 8\\ \end{bmatrix}