Show that $A_\mathrm{LBA}$ is PSPACE-Complete? I want to show that $A_\mathrm{LBA}$ is PSPACE-Compelte. 

Say we proved it is in PSPACE. Now for PSPACE-HARD:

I had an idea, which was very similar to some solution i found on the web- say we have a TM M which decides some language A in PSPACE. Then given an x in A we can simulate M on x to determine the number of steps M makes.

Suppose the above simulation is defined as another TM M' and k steps are needed for M to run x.

Build a TM M'' which simulates M on "x, $space^{\mathrm{k}}$" - the word x concatenated with k spaces. return (M'', "x, $space^{\mathrm{k}}$")

What i don't understand - the input is only x and its length is |x| so how do we know that simulating M on x is poly-time(x) ? we don't know the size of the description of M, which must be fed to the single-tape LBA M''. 

How come all the solutions i've encountered don't even mention that the size of the description of M might not be poly-time(x) ? then we have that the mere writing of M on M''s tape might be more than poly-time(x). (?)

No relation is given between |x| and |M| (the decider for A) because the input is only the word x...
As I understand, you are asking two questions. Let me address them separately: 


You don't. Your assumption on $M$ did not require anything more than $L(M) = A$, which is too weak for such a proof. You need to use $A \in \textbf{PSPACE}$ and require the existence of a polynomial $p:  \mathbb{N} \to \mathbb{N}$ such that $M$ uses at most space $p(n)$ for any input of length $n$.Incidentally, you also need this to know what $k$ should be. In your proof, you give no explicit way of computing $k$ other than simulating $M'$, but you do not state why this could ever be done in poly-time. (In fact, this could only ever work for arbitrary $M$ if $\textbf{P} = \textbf{PSPACE}$.) 
The description of $M$ remains constant as the length of the input $x$ increases, so it is negligible, asymptotically speaking. 




The real issue

It appears your approach is based on fixing $A$ and $x$ and then constructing an LBA $M''$ to simulate $M$ on $x$. This is not how reductions work. $A$ and $M$ should be fixed, but you need to keep $x$ arbitrary, otherwise the proof is quite incorrect (since the result of running $M$ on $x$ could be simply hardcoded into a TM which produces the result immediately upon reading $x$ as input; this proves absolutely nothing).