How is $a_j'= \sum\limits_{i=1}^n P_{ij}a_i$ shouldn't it be $a_i = \sum\limits_{j=1}^n P_{ij}a_j'$
I tried my best to understand how this is true and I spent more than an hour here
another question is what is $k$ in $a_k$ and what does it even show ?
this proof took me a lot of time and I couldn't figure it out.
They are correct. This is the standard change of basis machinery. It is perhaps unfortunate, however, that they use the letter $\alpha$ for the general vector in $V$ and also write $\alpha_1,\dots,\alpha_n$ (and $\alpha'_1,\dots,\alpha'_n$) for bases.
The columns of $P$ give the coefficients of the primed basis expressed as linear combinations of the unprimed basis. They then show that the columns of $Q=P^{-1}$ give the coefficients of the unprimed basis expressed as linear combinations of the primed basis. It often confuses students at the beginning, but the coordinates of a vector transform the opposite of the way the basis vectors transform: The vector of coefficients of an arbitrary vector in the primed basis is given by multiplying $P^{-1}$ (not $P$) by the vector of coefficients in the unprimed basis.
It helps to keep the following fundamental principle in mind: When you multiply a matrix $A$ by a vector $x$, the answer is a linear combination (with coefficients $x_j$) of the columns of $A$. At any rate, their computaton shows that the $k$th vector $\alpha_k$ of the original basis is obtained by taking the linear combination of the vectors $\alpha'_j$ given by the $k$th column of the matrix $Q$.
If you cannot understand their proof, you might want to look at my YouTube video on change of basis.