If $A=PDP^T$, does $P$ have to be orthogonal. A matrix A is called orthogonally diagonalizable if $A=PDP^{-1}$ and $A=PDP^{T}$, where $D$ is diagonal. Therefore, $P^{-1}=P^T$ and thus $P$ is an orthogonal matrix. If you are only given the fact that $A=PDP^T$, and $D$ is diagonal, is it guaranteed that $P$ is orthogonal and $A=PDP^{-1}$ also?
 A: 
A matrix A is called orthogonally diagonalizable if $A=PDP^{-1}$ and $A=PDP^{T}$, where $D$ is diagonal. Therefore, $P^{-1}=P^T$ and thus $P$ is an orthogonal matrix.

No, this is not true. E.g. when $D=0$, you have $PDP^{-1}=PDP^T$ for every invertible matrix $P$, but clearly not every invertible $P$ is orthogonal.
Your definition of orthogonal diagonalisation is different from the usual one. In the usual definition, $A$ is said to be orthogonally diagonalisable if $A=PDP^{-1}$ for some orthogonal matrix $P$. The definition demands that $P^{-1}=P^T$. It does not infer that $P^{-1}=P^T$ as a consequence.

If you are only given the fact that $A=PDP^T$, and $D$ is diagonal, is it guaranteed that $P$ is orthogonal and $A=PDP^{-1}$ also?

No. Consider $D=I$ and $P=\operatorname{diag}(1,2)$. This $P$ is not orthogonal and $PDP^T=\operatorname{diag}(1,4)\ne I=PDP^{-1}$.
A: No, your assertion is not true. Choose $$
 P = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix},
$$ and let $$D = I_2 = \begin{pmatrix}
1 & 0 \\ 0 & 1 
\end{pmatrix}$$ be the identity matrix (which is obviously diagonal) and $A$ defined by $$A = PDP^T = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix} \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix}^T = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix} \begin{pmatrix}
1 & 0 \\ 1 & 1
\end{pmatrix} = \begin{pmatrix}
2 & 1 \\ 1 & 1
\end{pmatrix}.
$$
But $$P^{-1} = \begin{pmatrix}
1 & -1 \\ 0 & 1 
\end{pmatrix} \neq P^T,$$
so $P$ is not orthogonal, and $$
PDP^{-1} = PP^{-1} = I_2 \neq A.
$$
A: No, for that to be true $A$ has to be a normal matrix. A normal matrix is such that
$$ AA^{*} - A^{*}A =  0  \tag{1}$$
e.g. 
$$ PDP^{T}(PDP^{T})^{T} - (PDP^{T})^{T}(PDP^{T})  = 0  \tag{2}$$
if they are orthogonal
$$ PDP^{T}PD^{T}P^{T} - PD^{T}P^{T}PDP^{T} = 0 \tag{3} $$
the $P$'s go away
$$ PD^{T}DP^{T}  - P D^{T}D P^{T} = 0 \tag{4} $$
