Let $(G_B)_{ij}=\langle b_i|b_j\rangle$ and $P_{BD}$ be the change of basis matrix, I'm asked to proof $$G_D=P^T_{BD}G_BP_{BD}.$$
I have tried to figure out why this works by just plugging in $\langle x|y\rangle$. It makes sense that we have $|y\rangle_B=P_{BD}|y\rangle_D$ as the ket, but I don't understand why the bra is $\langle x|_D=[x]^T_DP^T_{BD}G_B$.
Would it be correct to say $[x]_D=P_{DB}[x]_B \implies [x]^T_D=[x]^T_BP^T_{DB}$, after which we take right inverse of $P^T_{DB}$ to find $[x]^T_B=[x]^T_D(P^T_{DB})^{-1} = [x]^T_DP^T_{BD}$ so that substitution in $\langle x|_D=\langle x|_BG_B$ gives us the required bra?
Try to first understand why this is true in an $n$-dimensional vector space $V$. Choosing good notation helps. Let $E = [v_1, \dots, v_n]$ and $F = [w_1, \dots, w_n]$ be two ordered bases of $V$. Let $\langle \, , \, \rangle$ denote a symmetric bilinear form on $V$. Let $A = (a_{ij})$, where $a_{ij} = \langle v_i, v_j \rangle$; and let $B = (b_{ij})$, where $b_{ij} = \langle w_i, w_j \rangle$. Finally, let $P = (p_{ij})$ be the change of basis matrix such that $F = EP$, meaning that we regard $F$ and $E$ as row vectors of vectors. To be clear, this means that for each $j$, $w_j = \sum_{i=1}^{n} v_i p_{ij}$.
Now try to prove directly that $B = P^T A P$. To do so, simply compute $B$, replacing each occurrence of $w_j$ with $\sum_{i=1}^{n} v_i p_{ij}$. Use the fact that the form is bilinear.
The same argument applies in your geometric setting.