When is A a subset of the power set p(A)? I can across a video (here's the pic: [Discrete Math 1] Subsets and Power Sets) and the author mentioned the following:


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*Is A ⊆ p(A) for any A? No. 

*Is A ∈ (A) for any A? Yes.


p(A) is the power set of A in this example. I agree with these two points mentioned. 
However, I am confused with the exception he pointed out which is when p(∅). So we have p(A)={∅} and A=∅. It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).  
Is this because of ∅ being a subset of every set? If I go by this definition of a subset "The set A is a subset of the set B if and only if every element of A is also an element of B.", it does seem like this is one of the exceptions.
Is p({∅}) also an exception? A = {∅}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?
Here's a picture from the video: [Discrete Math 1] Subsets and Power Sets
 A: If $A=\emptyset$, then $A\subset P(A)$ is true, because $\emptyset\subseteq\{\emptyset\}$ is true. The latter is true because $\emptyset\subseteq B$ is true for any set $B$.
Your statement, however, is confusing:

It does look like A has an element ∅ that is in p(A), but it doesn't seem to be a subset p(∅).

The first part is not true. $A$ does not have an element. $\emptyset$ is not an element of $A$, because $A$ is the empty set, so $A$ has no elements. And precisely because $A$ has no element, every statement about "every element of $A$" must be true. In particular, the statement "every element of $A$ is an element of $P(A)$" is also true, and since that statement is the definition of $A\subseteq P(A)$, we see that $A\subseteq P(A)$ is also true.

Also, if $A=\{\emptyset\}$, then $P(A) = \{\emptyset, \{\emptyset\}\}$. In this case, again, $A\subseteq P(A)$, because every element of $A$ (there is only one, the empty set) is also an element of $P(A)$.

Even more generally, taking $A=P(P(\cdots P(\emptyset)\cdots))$, no matter how many powersets we take, we will always have $A\subseteq P(A)$.

EDIT:
I don't understand the video capture that you posted, but I will comment on this part:

Is p({∅}) also an exception? A = {{∅}}, so p(A) = {∅,{∅}}. Both sets contain ∅ as an element, so A here is a subset of p(A). Is this correct?

NO. You are not correct. If $A=\{\{\emptyset\}\}$, then $P(A)\neq \{\emptyset,\{\emptyset\}\}$. Remember, $P(A)$ is the set of all subsets of $A$. $\{\emptyset\}$ is not a subset of $A$, because it contains an element, $\emptyset$, which is not an element of $A$.
The correct answer with $A=\{\{\emptyset\}\}$ is this:
For $A=\{\{\emptyset\}\}$, the statement $A\subseteq P(A)$ does not hold, however. This is because, for $A=\{\{\emptyset\}\}$, $P(A)=\{\emptyset,\{\{\emptyset\}\}\}$. Now, there exists an element of $A$, in particular, $\{\emptyset\}$, which is not an element of $P(A)$.
A: We have $A\subseteq  p(A)$ if every element of $A$ is also a subset of $A$. That is if $x\in y\in A$ implies that also $x\in A$. Such sets aree called transitive. The most famous examples of transitive sets are the well-ordered transitive sets, aka. the ordinals. The exceptions you already noted are just the smallest ordinals, $\emptyset=:0$ and $\{\emptyset\}=:1$. 
