Two statements are equivalent if... Two statements "If A, then B." and "If P, then R." are equivalent if the validity of one statement implies the validity of the other?
That is, I assume that "If A, then B." is valid, then proceed to show the second statement is valid, and vice versa?
Or this done differently?
To show that a statement is valid, one need only show that if the premise is true, the conclusion necessarily follows, right?
 A: 
Two statements "If A, then B." and "If P, then R." are equivalent if the validity of one statement implies the validity of the other?

Not quite right.  It's not the the validity of the one statement implies the other (and vice versa), but rather that the assumed truth of the one implies the other (and vice versa)

That is, I assume that "If A, then B." is valid, then proceed to show the second statement is valid, and vice versa?

OK, so here's an example of what I mean.  By itself, $A \to B$ is not valid. So, anything would follow once you assume it is valid, which includes something like the validity of $C$. That is, by your proposed definition, the validity of $C$ folows from the validity of $A \to B$. But, clearly $C$ is not implied by $A \to B$, let alone that it would be equivalent to it.
So, you have to think about this in terms of the truth of the statements, rather than their validity: if you assume the $A \to B$ is true, does it mean that $C$ has to be true as well? No. So, $A \to B$ does not imply $C$ ... which is of course exactly what we want.
If you do want to talk about validity, you can do the following:
Two statements $\phi$ and $\psi$ are equivalent if and only if the inference from $\phi$ to $\psi$ is logically valid, and vice versa. Or: if the statements $\phi \to \psi$ and $\psi \to \phi$ are both valid.
A: Material equivalence between statements $p$ and $q$, denoted by  $p \leftrightarrow q$, is logically equivlaent to $(p \rightarrow q) \wedge (q \rightarrow p)$. 
Hence, if you wish to show that statements $a \rightarrow b$ and $p \rightarrow r$ are equivalent, or $(a \rightarrow b) \leftrightarrow (p \rightarrow r)$, then you may do this via natural deduction if you can show $[(a \rightarrow b) \rightarrow (p \rightarrow r)] \wedge [(p \rightarrow r) \rightarrow (a \rightarrow b)]$.
A general strategy to do this, known as a direct proof, would involve exactly what you are saying. First, assume $a \rightarrow b$ and then validly deduce $p \rightarrow r$. Secondly, assume $p \rightarrow r$ and then validly deduce $a \rightarrow b$. If you can do both, then you have demonstrated equivalence.
A: See at Archive.org : Seymour Lipschutz, Schaum's outline of set theory and relatid topics, chapter on Logic ( at the end). This book explains clearly the distinction between (1) material/ logical implication, (2) material biimplication / logical biimplication . Logical equivalence in precisely logical biimplication. It is a metalogical concept. 

You seem to presuppose that "being equivalent" is a relation that only holds betwen valid formulas. 
And that may come from the fact that " equi-valent" could be understood as " valid in the say way" or " having an equal validity". But this is misleading. 
In fact, the " equivalence" relation holds between any two formulas ( be they valid or not) as soon as these two formulas have the same truth value in all possible cases. 
Of course, all valid formulas ( being true in all possible cases) are equivalent ( i.e. have the same truth value - namely " True" - in all possible cases). 
But not all equivalent formulas are valid. For example : (A&B) and (B&A) are equivalent ( but far from valid). 
To show that two formulas X and Y are equivalent , one can : 
(1) show , using a truth table, that the material biconditionnal (X <-> Y) is valid 
Rk : if X = (A--> B) and Y = (P--> R), this method amounts to showing that the whole biconditional formula : [ (A-->B) <--> (P--> R) ] is valid ( i.e. is a tautology) 
OR 
(2) show that the two material implications (X --> Y) and (Y --> X) are both valid , that is , that X and Y both logically imply each other 
Rk : this is not the same thing as showing that X by itself is valid, that is to say that (A --> B) by itself is valid, in case X happens to be defined as (A-->B) as it is in your example ; nor is it the same thing as showing that Y by itself is valid; what is to be shown is that the whole big conditionals (X -->Y ) on one side and (Y --> X) on the other side are valid; *the contingent fact that in your example X and Y are already (small) conditionals musn't occasion a confusion in your mind.* 
(3) without any premise, assume X ( for conditional proof) and derive Y to show that (X --> Y) is valid ; then  assume Y to derive X to show that (Y--> X) is valid, then finally use &-intro to conclude that (X<-->Y) is valid. 
