Two $n \times n$ matrices $A,B$ are said to be similar if there exists a non-singular matrix $P$ such that ${P^{ - 1}}AP = B$
If $A$ and $B$ are two non-singular matrices, then
A. $A$ is similar to $B$
B. $AB$ is similar to $BA$
C. $AB$ is similar to ${A^{ - 1}}B$
D. none of these.

Hint: For the similar matrices $A,B$ we can write that ${P^{ - 1}}AP = B$ or vice versa where $P$ is given $n \times n$ invertible matrix. And it is also given that $A,B$ are the non-singular matrix that means that the determinant of $A{\text{ and }}B$ cannot be zero. So as we know that ${P^{ - 1}}AP = B$ where $P$ is any matrix. So $A{\text{ or }}B$ can be taken in place of$P$. We know that ${A^{ - 1}}A = I$

Complete step-by-step answer:
Here we are given that $A{\text{ and }}B$ are two $n \times n$square matrices and also it is given that they both are similar. As we know that whenever we are given that two matrix like $A{\text{ and }}B$are similar then
 we can write that ${P^{ - 1}}AP = B$ or we can also write that ${P^{ - 1}}BP = A$
Here $P$ is any $n \times n$ square matrix. So as we also know that $A{\text{ and }}B$ are two $n \times n$ square matrix.
Also we are given that $A{\text{ and }}B$ are non-singular that means that the determinant of $A{\text{ and }}B$ cannot be zero. So we must know that the product of any matrix with its inverse gives the identity matrix.
So we can write that ${A^{ - 1}}A = I$ and also ${B^{ - 1}}B = I$
Here $A{\text{ and }}B$ are two $n \times n$ square matrix and $I$ is the identity matrix.
So as we can write that
$AB = AB$as here LHS is equal to the RHS
Now we know that if we multiply any matrix by the identity matrix gives us the same matrix that means $AI = A,BI = B$
So we can multiply by $I$ in RHS
$AB = IAB$
Now we can write ${B^{ - 1}}B = I$
So putting the value we get
$AB = ({B^{ - 1}}B)AB$$ - - - - - (1)$
Now we know that multiplication of the matrix follow the associative properties that means if $A,B,C$ are three matrices then we can write that $(AB)C = A(BC)$
So using this property in the equation (1)
$AB = {B^{ - 1}}(BA)B$
Now if we know that
${P^{ - 1}}AP = B$, then $B$ is similar to $A$
Where $P$ is any $n \times n$ matrix and here we are given that
$AB = ({B^{ - 1}}B)AB$ where $B$ is any $n \times n$ matrix, so we can say that $AB,BA$ are similar matrices.

Note: We know for any matrix the product of that matrix with its inverse gives us the identity matrix which is denoted by$I$. $I$ can be $n \times n$ matrix which is denoted by ${I_n}$
For example: ${I_2} = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$ and ${I_3} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
And so on.
So we can write that $A.{A^{ - 1}} = I$