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I am asked to show that if $A\cap B = \{ \phi \},$ then $ P(A) \leq \overline{P(B)}$

I understand that my first step is to recognize that $A$ and $B$ are disjoint, since their intersection is the empty set. The solution that I am referencing from goes on to complete the explanation with the following: enter image description here

I do not understand why we can say $A$ is a subset of $B^C$, and because of that, I am confused as to how the remainder of the solution is made. I am having a tough time pulling these relationships from memory, and was also wondering if anyone has any tricks for understanding it better.

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    First, the symbol is $\emptyset$, not $\phi$ (``\emptyset`` in TeX). Also, you mean $A\cap B=\emptyset$, not $\{\emptyset\}$. The latter is **the set containing the empty set**, which is not equal to the empty set. I think there are some other issues with the question too. – 2017-01-18

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As parsiad commented, "$A \cap B = \{\emptyset\}$" is a different statement from "$A$ and $B$ are disjoint". I will assume you really mean they are disjoint, since that's the only way the question makes sense.

If $A$ is not a subset of $B^C$, then there is some element $x \in A$ not contained in $B^C$. What can you say about the relationship between $x$ and $B$? What does this imply about $A$ and $B$?

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Draw a picture. If $A$ and $B$ do not intersect, then every element of $A$ is not an element of $B$, and thus every element of $A$ belongs to $B$'s complement.