(in the question indexed $\mathbf{E}$ are contravariant basis vectors of a given coordinate system)
The equation in the title of the question is a consequence of the solution of this exercise (I introduced $p$ and $q$ to not repeat the same symbol for different dummy index in the same equation, but the tensors $\mathbf{A}$ and $\mathbf{B}$ are the same in both side of the equation)
I could ask the question like this: tensor calculus books are said to pay carefully attention to the index order; but when the inner product is made, the indexes can be moved back and forth? In the first solution $\mathbf{E}_k$ is pushed behind $\mathbf{E}_j$, is it permissible (in the second solution too $\mathbf{E}_k$ is pushed back but it is different, after marrying $\mathbf{E}_j$ whit inner product, the object loses all its free index and it became a scalar)? If this is the case, if it is doable, how could be true that, for higher rank tensor, inner product is not commutative? If both solution are correct and the equation I wrote in the title works, how can I see that the two vectors obtained are the same? I'm confused in understanding the permissible ways in doing inner product. If only I had a simple geometrical meaning like vectors inner products, probably all would be clear. As far as I see, a rank 2 tensor in a $n$-dimension space is a sort of linear combination of $n^2$ dyads (and each dyad is a ordered double vector: we make $n^2$ dyads coupling in all the way a vector basis of the space) like a vector is a linear combination of $n$ vector. Is this correct? If this is the case, in this perspective, is there a simple way to see geometrically what a tensor inner product is? If answering to all this is too long I'll understand things later, but please simply tell me if moving $\mathbf{E}_k$ as the book do is correct and so if works $A_{ij} B^i \mathbf{E}^j = A_{pq} B^q \mathbf{E}^p$.
Edit
I expanded. $\mathbf{E}^i$ are linearly independent so the only way the equation works is that works $$A_{11}B^1+A_{21}B^2+...+A_{n1}B^n=A_{11}B^1+A_{12}B^2+...+A_{1n}B^n$$ and other $n$ similar equations. Simmetry of $\mathbf{A}$ would ensure that, but in general this is not true. The book is Taha Sochi's one (solutions, pag 75).
$$ A_{ij}B^i {\bf E}^j = A_{qp} B^q {\bf E}^p $$ since one can just change the names of dummy indices. In general, $A_{qp} \ne A_{pq}$ so the result is not equal to $A_{pq} B^q {\bf E}^p$. However, if $A$ is a symmetric tensor then $A_{qp} = A_{pq}$ and then the equation you asked about does hold.