There is a canonical bijective correspondence between the isometry classes of symmetric bilinear spaces and congruence classes of symmetric matrices.
I know that we need to show every σ which belongs to isometry class such that b(σx,σy)=b(x,y) has one-one correspondence with congruence classes of matrix B which is matrix of bilinear form and its congruence classes looks like $S^T$BS where S is associated change of basis matrix. My question is what it means by finding canonical map between isometry classes and congruence classes and how that map gives bijective correspondence? Bilinear form b is of form b(x,y)= $x^T$By. b($\sigma$x,$\sigma$y)=$x^T$SB$S^T$y
Given a class of isometric symmetric bilinear space $[(V,b)]$ a class of congruent symmetric matrix can be assigned in this way.
Take a representative of $[(V,b)]$. Let it be $(V, b)$. Given a vector basis $\mathfrak{E}(V)$ of $V$ and an isometry of $(V, b)$ on $K^n$, $b$ is represented by a symmetric matrix $B$. Choosing any other vector basis and/or any other isometry of $(V,b)$ on $K^n$ the symmetric matrix representing $b$ is congruent to $B$. Moreover the converse is also true, that is, any matrix congruent to $B$ represents $b$ under some isometry and vector basis. Let's call $[B]$ the class of matrices conguent to $B$.
Then from $(V,b)$ we have arrived at $[B]$. If you choose any other element of $[(V,b)]$ we would arrive at the same $[B]$.
$[(V,b)]\mapsto [B]$ is bijective. Indeed, let $(W, c)\notin[(V,b)]$, whatever basis is chosen in $W$ and whatever isometry of $(W,c)$ onto $K^n$ is chosen the symmetric matrix representing $c$ would never be $C\in[B]$. And of course it is surjective.
Alternatively, if you want you could also say that there exist symmetric matrices $B$ such that $(V,b)\cong\langle B\rangle$ then the wanted bijection is $[\langle B\rangle] \mapsto[B]$ (where $\langle B\rangle=(K^n,b_B)$ and $b_B(x,y)=x^tBy$)