Reducing one problem to another

I know this is sort of a basic question, but I don't completely understand the following. Let $$A$$ and $$B$$ be two problems. If I take one instance $$a$$ of $$A$$ and one instance $$b$$ of $$B$$, and show the following. Solving $$a$$ implies solving $$b$$, and solving $$b$$ implies solving $$a$$. Am I showing that $$A$$ reduces to $$B$$, $$B$$ reduces to $$A$$ or both? How are $$A$$ and $$B$$ related using the $$\leq_P$$ and $$=_P$$ notation?

Neither -- you can't show a reduction by starting out with an instance of $$A$$ and an instance of $$B$$. Instead, a reduction has to be a way of transforming instances of $$A$$ into instances of $$B$$, and vice versa.
A reduction from $$A$$ to $$B$$ means you give a way of transforming any instance $$a$$ of $$A$$ to an instance $$f(a)$$ of $$B$$, such that $$a$$ is a yes-instance of $$A$$ if and only if $$f(a)$$ is a yes-instance of $$B$$. Here $$f$$ is the reduction, or transformation. Notice that $$f(a) = b$$ might not just be any instance of $$B$$, but it might be a very particular instance constructed by our transformation.
For example: suppose $$A$$ is the problem of determining if a number is even, and $$B$$ is the problem of determining if a number is a multiple of 10. Then to give a reduction from $$A$$ to $$B$$ we could start with a number $$n$$, and multiply it by $$5$$. Our reduction is:
This is a valid reduction because n is an even if and only if n * 5 is a multiple of 10. But, it doesn't construct every possible instance of $$B$$; only multiples of $$5$$ are actually constructed as instances of $$B$$, not all integers.