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.

This one $$\{\varnothing\}\subseteq\{\{\varnothing\}\}$$ is incorrect.
$A\subseteq B$ means that for every $x\in A$ it is true that $x\in B$.
$\{\varnothing\}$ contains one element: $\varnothing$, but $\{\{\varnothing\}\}$ doen not contain $\varnothing$ so $$\{\varnothing\}\not\subseteq\{\{\varnothing\}\}$$
The other two are correct. $$\mathcal{P}(\{\{\varnothing\}\})=\{\varnothing,\{\{\varnothing\}\}\}$$
And as always: power set of set which contains one element contains two elements.
