$f:A\to B$ means $f$ is a function from $A$ to $B$.
Example:
$\begin{align*}f:\Bbb R& \to \Bbb R_+\\
x & \mapsto x^2\end{align*}$
You've certainly already seen functions defined as $f(x)=x^2$ but as you start doing more complicated things with functions, you need the "formula" plus two other things: the domain $A$ and the codomain $B$.
The reason fo that is that the function I defined above is not bijective (if I give you $f(x)$, you can not find $x$ because it could be $\sqrt{f(x)}$ or $-\sqrt{f(x)}$) but I can define another function that is bijective:
$\begin{align}g:\Bbb R_+& \to\Bbb R_+\\
x & \mapsto x^2\end{align}$
Because now you know the $x$ I took to form $f(x)$ is in $\Bbb R_+$ so it can not be $-\sqrt{f(x)}$ so it has to be $\sqrt{f(x)}$.
Therefore you can define
$h:\begin{array}{ll}\Bbb R_+& \to& \Bbb R_+\\
x & \mapsto & \sqrt{x}\end{array}$
And $h$ will be the inverse function of $g$ which we write as $g^{-1}=h$. Also note that $f$ does not have an inverse function.
Given any monoid (that is a set $A$ equipped with an associative operation $\cdot$ and an identity $1$), we can define "finite products" roughly by:
$$\prod_{i=1}^n a_i = a_1\cdot a_2\cdot \dots \cdot a_n$$
where $\prod_{i=1}^0 = 1$.
Possible monoids are for example $(\mathbb{R},\cdot, 1)$ yielding "$\prod$", $(\mathbb{R},+,0)$ yielding "$\sum$" or $(P(S), \cup, \emptyset)$ yielding "$\bigcup$" and so on and so forth.
So, we can also view a monoid as a set $A$ together with a map $A^* \to A, (a_n) \mapsto \prod_{i=1}^n a_i$ taking lists (words, tuples) of elements of $A$ to elements of $A$.
Occassionally however, we may find "maps" (broadly speaking) which not only accept finite lists, but also infinite lists or even bigger families of elements as objects.
For example, a complete lattice is a set $A$ equipped with maps $\bigvee$ and $\bigwedge$ taking abitrary families of elements of $A$ to elements of $A$.
Intuitively, if you take the set of all "small sets" (this is usually realized as a proper class) as the set $A$, then you get a complete lattice with operations $\bigcup$ and $\bigcap$ called union and intersection, which take families of elements of $A$ (that is sets of sets) to sets.
Best Answer
The triangle is almost certainly being used to for symmetric difference:
$$A\mathrel{\triangle}B=(A\setminus B)\cup(B\setminus A)=(A\cup B)\setminus(A\cap B)\;.$$
The $P$ might be the power set operator, in which case the whole thing is the set of all subsets of $A\mathrel{\triangle}B$:
$$P(A\mathrel{\triangle}B)=\{X:X\subseteq A\mathrel{\triangle}B\}.$$
However, a script $P$ is more often used for this purpose, and as has just been pointed out to me, you used the probability tag, so $P(A\mathrel{\triangle}B)$ is probably the probability assigned to the subset $A\mathrel{\triangle}B$ of your sample space.