If $P$ be permutation matrix then $({P^T}AP)x\mathop = \limits^? \left( {\begin{array}{*{20}{c}} * \\ 0 \\ \end{array}} \right)$ 
*

*Let $A\in M_n$ is nonnegative(all $a_{ij}\ge0$).

*$x\in C^n$  be eigenvector of $A$ with $r ≥ 1$ positive entries and $n − r$ zero entries .

*There is $P\in M_n$(permutation matrix) such that  $Px = \left( {\begin{array}{*{20}{c}}
   {x'}  \\
   0  \\
\end{array}} \right)$ where $x' > 0$.


Can we say that $({P^T}AP)x\mathop  = \limits^? \left( {\begin{array}{*{20}{c}}
   *  \\
   0  \\
\end{array}} \right)$
.

No, this won't hold for general matrices $A$.
For instance, you can take $n=2$, and $x = \left( {\begin{array}{*{20}{c}}
   0  \\
   1  \\
\end{array}} \right)$. Then the permutation matrix you are looking for is $P = \left( {\begin{array}{*{c}{c}}
   0 & 1 \\
   1  & 0\\
\end{array}} \right)$, and $Px = \left( {\begin{array}{*{20}{c}}
   1  \\
   0  \\
\end{array}} \right).$
Now take $A = \left( {\begin{array}{*{c}{c}}
   1 & 1 \\
   1  & 1\\
\end{array}} \right)$. Then $APx = A(Px) = \left( {\begin{array}{*{20}{c}}
   1  \\
   1  \\
\end{array}} \right)$, and so $(P^TAP)x = P^T(APx) = \left( {\begin{array}{*{20}{c}}
   1  \\
   1  \\
\end{array}} \right)$.
