Can the cardinality of a power set ever be odd? Can the cardinality of a power set ever be odd? If it can, what conditions must be met?
 A: The power set of the empty set has 1 element, the empty set. $2^0=1$
EDIT: Credit as well to Tunococ, Andreas Caranti, and Mufasa for their comments. I guess I jumped the gun since this is my first answer!
A: No, this is not possible for finite non empty sets. This is because you have $2$ possibles "truth values" : true or false.
More precisely, let $E$ a finite set with $n≥1$. Then the power set has $2^n$ elements which is even if $n≠0$.
Let $A \subseteq E$. For each $x \in E$, either $x\in A$ or $x \not \in A$. In other words, for each of the $n$ elements, you have $2$ possibilites : either it belongs to $A$ or it doesn't.
Therefore you have 
$\underbrace{2 \cdot 2 \cdots 2 \cdot 2}_{n \text{ times}} = 2^n$
 different possibilities, corresponding to the $2^n$ different possible subsets of $E$.
A: To summarize some of the previous answers and comments here in mathematical notation:

Can the cardinality of a power set ever be odd?

$S=\emptyset \implies P(S)=\{\emptyset\} \implies |P(S)|=1 \implies |P(S)|\equiv1\pmod2$

If it can, what conditions must be met?

$|P(S)|\equiv1\pmod2 \implies |P(S)|=1 \implies P(S)=\{\emptyset\} \implies S=\emptyset$
