From Wikipedia:
The empty set $\emptyset$ has exactly one partition, namely $\emptyset$.
I believe this example to be wrong. A partition must have, by definition, nonempty cells, so $P = \{ \emptyset \}$ is not allowed. If I am reading correctly the definition of $P$, the empty set admits no partition.
On
It's not $\{\emptyset\}$, but $\emptyset$. That is, $P$ (the partition itself) has no elements. This means:
Every element of $P$ is nonempty (since it doesn't have any).
Meanwhile, the union of the elements of $P$ is $\emptyset$: every element of $\emptyset$ is in some element of $P$ (since $\emptyset$ doesn't have any elements), and every element of an element of $P$ is an element of $\emptyset$ (since $P$ doesn't have any elements).
So $P$ is indeed a partition of $\emptyset$, albeit in a silly way.
On
Here is another way of seeing this: recall that $P$ is a partition of $A$ if and only if there is an equivalence relation $E$, such that $P=A/E$.
If $A=\varnothing$, then $A/E$ has got to be the empty set, since there is a surjection from $A$ onto $A/E$. Therefore, $A/E=\varnothing$, so the only partition of $\varnothing$ is $\varnothing$.
Yes, the empty set does have a partition. Let's see what a partition is: given a set $X$, a partition of $X$ is a set $P$ of nonempty subsets of $X$ such that each element of $X$ is contained in exactly one element of $P$.
Consider $P = \varnothing$.
Is it a collection of nonempty subsets of $\varnothing$? Yes, all the elements of $P$ are nonempty subsets of $\varnothing$, because $P$ has no elements so the assertion is vacuously true.
Is every element of $\varnothing$ contained in exactly one element of $P$? Again yes, this is vacuously true.
Therefore $P = \varnothing$ is indeed a partition of $\varnothing$. However $P = \{ \varnothing \}$, the set with one member, is not a partition of $\varnothing$, because it fails the first requirement.