If A is symmetric matrix,

It is given that, $A$ is symmetric matrix.

$\therefore A^{T}=A$            ...(1)

Now,

$\left(B^{T} A B\right)^{T}$

$\left(B^{T} A B\right)^{T}$

$=B^{T} A^{T}\left(B^{T}\right)^{T} \quad\left[\right.$ For any matrices $\left.X, Y, Z,(X Y Z)^{\top}=Z^{\top} Y^{\top} X^{\top}\right]$

$=B^{T} A B$      [Using (1)]

Since $\left(B^{T} A B\right)^{T}=B^{T} A B$, so the matrix $B^{\top} A B$ is symmetric.

If $A$ is symmetric matrix, then $B^{\top} A B$ is symmetric