@copper.hat If $B$ is countable we take a bijection between $B$ and $\mathbb{N}$.
If $B$ is uncountable, it has a countable subset $B'$. We take a bijection between $B'$ and $\mathbb{N}$.
Then we take a bijection between $\mathbb{N}$ and $\mathbb{Q}$.
We define the set $\Gamma_r=\{x \in \mathbb{Q} \mid x<r\}, r \in \mathbb{R}$.
This set defines an injective order-preserving map from $\mathbb{R}$ to $\mathcal{P}(\mathbb{Q})$.
$\{\Gamma_r\}$ is a chain because if $r<s$ then $\{x\in\Bbb Q:x<r\}\subset\{ x\in\Bbb Q:x<s\}$ and it is uncountable because of the following:

Let $B$ be an infinite set. It has an infinite countable subset $B'$.
Since $\mathbb{Q}$ in countable, we can take a bijection between $B'$ and $\mathbb{Q}$.
We define the set $\Gamma_r=\{x \in \mathbb{Q} \mid x<r\}, r \in \mathbb{R}$.
$\{\Gamma_r\}$ is a chain because if $r<s$ then $\{x\in\Bbb Q:x<r\}\subset\{ x\in\Bbb Q:x<s\}$ and it is uncountable because the $\Gamma_r$s are distinct and there are uncountably many reals.
We have a bijection between $B'$ and $\mathbb{Q}$, so there is also a bijection between $\mathcal{P}(B')$ and $\mathcal{P}(\mathbb{Q})$.