On the one hand I feel as if there is not enough information, as we don't get information on $P(b)$, which is what we would need to use the law of total probability.
But, on the other hand, $P(a)$ seems to be independent from $P(b)$ so maybe $P(a) =P(a|b)=P(a|b^c)$
On
By the law of total probability, $$P(A) = P(A|B)P(B)+P(A|B^c)P(B^c)= P(A|B)P(B)+P(A|B)(1-P(B))=P(A|B)$$
On
Yes, indeed, the equality of those conditional probability indicates that $A$ and $B$ are independent, and that they equal the marginal probability too.
$$\begin{align}\mathsf P(A\mid B)&=\mathsf P(A\mid B^\complement)&&\text{Premise}\\[2ex]\mathsf P(A)&=\mathsf P(A\mid B)~\mathsf P(B)+\mathsf P(A\mid B^\complement)~\mathsf P(B^\complement)&&\text{Law of Total Probability}\\[1ex]&=\mathsf P(A\mid B)~(\mathsf P(B)+\mathsf P(B^\complement))&&\text{Distribution & by premise}\\[1ex]&=\mathsf P(A\mid B)&&\text{Rule of Probability for Complements}\\[2ex]\mathsf P(A)~\mathsf P(B)&=\mathsf P(A\mid B)~\mathsf P(B)&&\text{Multiplication by same factor}\\[1ex]&=\mathsf P(A\cap B)&&\text{Definition of Conditional Probability}\\[3ex]\therefore \quad A&\perp B\end{align}$$
I would start by noticing that
\begin{align*} \textbf{P}(A|B) = \textbf{P}(A|B^{c}) & \Longleftrightarrow \frac{\textbf{P}(A\cap B)}{\textbf{P}(B)} = \frac{\textbf{P}(A\cap B^{c})}{\textbf{P}(B^{c})} = \frac{\textbf{P}(A) - \textbf{P}(A\cap B)}{1-\textbf{P}(B)}\\\\ & \Longleftrightarrow \textbf{P}(A\cap B) = \textbf{P}(A)\textbf{P}(B) \Longleftrightarrow \textbf{P}(A|B) = \textbf{P}(A) \end{align*} Based on the given information, this is as far as one can get, that is to say, to conclude that $A$ and $B$ are independent events.