Power set of $\{\{\varnothing\}\}$ $$\mathcal{P}(x)=\{y\mid y\subseteq x\}$$
$$\mathcal{P}(\varnothing)=\{\varnothing\}$$
$$\mathcal{P}(\{\varnothing\})=\{\varnothing,\{\varnothing\}\}$$
$$\mathcal{P}(\{a,b\})=\{\varnothing,\{a\},\{b\},\{a,b\}\}$$
For $\mathcal{P}(\{\{\varnothing\}\})$ we have: $$\varnothing\subseteq\{\{\varnothing\}\}$$
$$\{\varnothing\}\subseteq\{\{\varnothing\}\}\text{ and}$$
$$\{\{\varnothing\}\}\subseteq\{\{\varnothing\}\}$$
Therefore $\mathcal{P}(\{\{\varnothing\}\})=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\}\}$
Is this answer correct? Unsure with this topic, just need some verification, thanks.

What is the power set of a one-element set? Work out $\mathcal{P}(\{a\})$ for the singleton set and substitute $a = \{\varnothing\}$
The fact that $\{\varnothing\}$ is your set element (and is itself a set) is what appears to be confusing you. Forget what it is, it's just an element.
