Multiple Necessary and Sufficient Conditions Does if and only if allow for multiple necessary and sufficient conditions?
In other words, can there be multiple hypotheses that are both necessary and sufficient for a conclusion? Any examples would be great.
Also, can you provide any intuition as to why $p \leftrightarrow q \wedge p \leftrightarrow r$ is not logically equivalent to $p \leftrightarrow (q \wedge r)$?
 A: Your $p\leftrightarrow (q\land r)$ is an example of multiple necessary and sufficient conditions for $p$. 
Using a truth table, you can confirm that the following is not always true:
$$(((p\leftrightarrow q) \land  (p\leftrightarrow r))\leftrightarrow(p\leftrightarrow(q\land r)))$$
It is false only when $p$ is false and only one of $q$ and $r$ is true.
(Nice truth table generator at http://web.stanford.edu/class/cs103/tools/truth-table-tool/ A clean, slick interface.)
A: 
Also, can you provide any intuition as to why $p \leftrightarrow q \wedge p \leftrightarrow r$ is not logically equivalent to $p \leftrightarrow (q \wedge r)$?

It is because, if (and only if) two predicates are equivalent, then also their negations are equivalent.
$$\begin{align}A\leftrightarrow B \quad\equiv&\quad \neg A\leftrightarrow \neg B
\\ p \leftrightarrow q~\wedge ~p\leftrightarrow r  \quad\equiv&\quad \neg p\leftrightarrow\neg q~\wedge~\neg p\leftrightarrow \neg r
\\ p~\leftrightarrow~(q\wedge r) \quad\equiv&\quad \neg p~\leftrightarrow(\neg q\vee \neg r)
\end{align}$$
In the first case, $p\leftrightarrow q~\wedge~ p\leftrightarrow r$, then if $p$ is false so too are both $q$ and $r$ (ie: neither may be true).
In the second case, $p ~\leftrightarrow~ (q \wedge r)$, then if $p$ is false so too is either $q$ or $r$ (ie: at most one may be true). [vis: de Morgan's Laws]

$$\begin{align}(p\leftrightarrow q)~\wedge~(p\leftrightarrow r) \quad\equiv&\quad (p\to q)~\wedge~ (p\to r)~\wedge~\color{silver}{(}(q\to p)~\wedge~(r\to p)\color{silver}{)}\\p~\leftrightarrow~(q\wedge r) \quad\equiv&\quad (p\to q)~\wedge~ (p\to r)~\wedge~((q\to p)~\vee~(r\to p))\end{align}$$
