Prove or disprove : $P[A|B] = P[B]$, the A and B are independent? Is this right? SOrry if this is extremely easy.
I did the following but I'm a little bit unsure about it:

Let $A=B$, and $P[A]>0$.
Then $$P[A|A] = P[A]$$
But A is not independent with itself:
$$P[AA] = P[A] \neq P[A]^{2} = P[A]P[A] $$
Thus, the preposition does not hold.

I'd appreciate your comments on this one.
Thanks!

The error in the quoted reasoning is in  $P[A|A] = P[A]$. Instead we should have: $$P[A|A] = 1 \neq P[A]$$
For example, let the event be '$A = \text{it rains}$'.
Say you live in the desert where it almost never rains, but suppose that it is raining outside, then what is the probability that it is raining outside?

Edit: I notice now that you have to prove or disprove $P[A|B] = P[B]$. I misread this as $P[B|A] = P[B]$ which is true by definition when $A$ and $B$ are independent.

$P[A|B]=P[B]$ if the $A$ and $B$ are independent

This is not true. Counterexamples occur when $P[A] \neq P[B]$.
In case of independence you have by definition $P[A|B] = P[A]$ and that means that for any case where $P[A] \neq P[B]$ you have $$\rlap{\overbrace{\phantom{P[A|B] = P[A]}}^{\text{by definition}}} P[A|B] = \underbrace{P[A]\neq P[B]}_{\text{by assumption}}$$ Thus $P[A|B] = P[B]$ does not (necessarily) follow from independence, it is only true when also $P[A] = P[B]$ and the statement fails in cases where $P[A] \neq P[B]$.

If $P[A|B]=P[B]$, then the $A$ and $B$ are independent

This is neither true.
Counterexample: Any case where dependent $A$ and $B$ are dependent but $P[A|B]=P[B]$. (alternatively one can use cases with $P[A] \neq P[B]$)
An example is the case where we toss a fair coin two times and define $B$ be the event 'heads on the first toss', and let $A$ be the event 'heads on both tosses'. Then the table of probabilities looks as following
$$\begin{array}{ccc}
& \text{A: heads both tosses} & \text{not A: no heads both tosses} \\
\text{B: heads first toss} & 0.25 & 0.25 \\
\text{not B: no heads first toss} & 0 & 0.5 \\
\end{array}
$$
From these we can deduce $$P[A|B] = \frac{P[\text{$A$ and $B$}]}{P[\text{$A$ and $B$}]+P[\text{(not $A$) and $B$}]} = \frac{0.25}{0.25+0.25} = 0.5$$
and
$$P[B] = P[\text{$A$ and $B$}]+P[\text{(not $A$) and $B$}] = 0.25 + 0.25 = 0.5$$
And in this case we have $P[A|B]=P[B]$ but $A$ and $B$ are dependent. Therefore the statement is proven false by contradiction.
The condition $P[A|B]=P[B]$ does not (generally) imply that $A$ and $B$ are independent.
