Let $A$ be the matrix of a bilinear form in a certain basis $\{v_1,v_2,v_3\}$, where $A$ is:
$$ A= \left[ \matrix { 1 & 0 & 2 \\ 0 & -1 & 1 \\ 2 & 1 & 1 } \right] $$
I've diagonalized it, and the resulting diagonal matrix is $D=diag(1,-1,-2)$, and the change of basis matrix is:
$$ P= \left[ \matrix { 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 1 & 1 } \right] $$
So, its verified that $D=PAP^t$. However, I'm not sure how to find in which base this happens, in terms of the vectors of the old base, of course.
I'm not sure how you found that $P$. In any case it seems you got the transposition on the wrong side: base change of bilinear forms has the transpose of the change of basis matrix (or of its inverse, depending on the direction) on the left. It is therefore the rows of you $P$ (interpreted on the old basis) that give an orthogonal basis for the bilinear form associated to your matrix$~A$.