What is the difference between using $PAP^{-1}$ and $PAP^{T}$ to diagonalize a matrix? What is the difference between using $PAP^{-1}$ and $PAP^{T}$ to diagonalize a matrix? 
Can both methods be used to diagonalize a diagonalizable matrix $A$?
Also does $A$ been symmetric or not effect which method to use?
 A: You always use $PAP^{-1}$ to diagonalize a matrix, since this process (multiplying by $P$ on one side and $P^{-1}$ on the other) is effectively changing the basis of your matrix, and so is not changing many of the fundamental properties.
If $P^{-1} = P^T$ then $P$ is called an orthogonal matrix, and $PAP^{-1}$ is the same thing as $PAP^T$. A matrix is orthogonally diagonalizable (diagonalizable by an orthogonal matrix) if and only if it is symmetric.
A: The difference is the following: When you use the formula $$P A P^{-1}$$ is because you regard the matrix $A$ as the matrix of a linear map $f$. Then the above formula tell you how the matrix $f$ change when you change the coordinates i.e. the basis. Instead the formula $$P A P^{\top} $$
is used when you regard $A$ as the matrix of a quadratic form $q$. Then the above formula tell you how change the matrix of $q$ when you change the coordinates i.e. the basis. 
So in order to perform a diagonalization you have to be aware if $A$ is the matrix of either a linear map or of a quadratic form. In the special that $P^{\top} = P^{-1}$ i.e. when $P$ is orthogonal both formulae agree. This happens in the traditional method of diagonalization of a symmetric matrix by means of it eigenvectors. But in order to diagonalize a quadratic form it is not necessary to use an orthogonal matrix $P$. Such diagonalization can be done by the traditional method of completing the square:
 https://en.wikipedia.org/wiki/Completing_the_square 
A: You can think of a matrix $A$ as either a linear operator ($x \mapsto Ax$) or a bilinear form ($x,y \mapsto <x,Ay>$). Diagonalizing it using $PAP^{-1}$ diagonalizes the linear operator $A$, i.e. changes the basis of the vector space to one in which the linear operator is diagonal, whereas diagonalizing it using $PAP^{T}$ diagonalizes the bilinear form, i.e. changes the basis of the vector space to one in which the bilinear form is diagonal.
