The axiom of choice implies that for every partial order $P$ the
hypergraph $H_P$ has property $B$.
Let $(P,\le)$ be a partial order.
We first claim the following: for every $p\in P$ there is a $q\le p$ such that
for all $r\le q$ one has $\bigl|d(r)\bigr|=\bigl|d(q)\bigr|$.
This follows from the obvious fact that $x\mapsto\bigl|d(x)\bigr|$ is a
monotone map from $P$ to the class of cardinal numbers plus that the negation
of our claim would yield a decreasing sequence $\langle x_n:n\in\omega\rangle$
such that $\bigl|d(x_n)\bigr|>\bigl|d(x_{n+1})\bigr|$ for all~$n$, which is
impossible as the cardinal numbers are well-ordered.
Let $Q$ be the set of elements of $P$ with the property in the claim.
So if $q\in Q$ then $\bigl|d(r)\bigr|=\bigl|d(q)\bigr|$ for all $r\le q$.
If $q\in Q$ then $d(q)=\{q\}$ or $d(q)$ is infinite.
Indeed $d(q)$ were finite and $r<q$ then $q\in d(q)\setminus d(r)$ and so $\bigl|d(r)\bigr|<\bigl|d(q)\bigr|$,
a contradiction.
Note also: if $q_1,q_2\in Q$ are such that $d(q_1)$ is infinite
and $d(q_2)=\{q_2\}$ then $d(q_1)$ and $d(q_2)$ are disjoint.
Let $A$ be a maximal antichain in $Q$ in the sense that the family
$\{d(q):q\in A\}$ is pairwise disjoint and that $A$ is maximal with this
property.
Divide $A$ into $A_1=\{q\in A:d(q)=\{q\}\}$ and its complement $A_2$.
If $q\in A_2$ then $d(q)$ is infinite and, as is well-known, one can construct a pairwise
disjoint family $\{e(r):r\le q\}$ such that $e(r)\subseteq d(r)$ and
$\bigl|e(r)\bigr|=\bigl|d(q)\bigr|$ for all $r\le q$.
This makes it easy to define $f_q:d(q)\to\{0,1\}$ such that $f_q$ is not
constant on any $d(r)$ with $r\le q$.
Then $f_2=\bigcup_{q\in A_2}f_q$ defines a function (colouring) on
$R=\bigcup_{q\in A_2}d(q)$.
Define $f_1:P\setminus R\to\{0,1\}$ as follows: if $q\in A_1$ then $f_1(q)=0$,
otherwise $f_1(p)=1$.
Then the map $f=f_1\cup f_2$ witnesses property $B$.
Let $p$ be such that $d(p)$ has at least two elements.
By maximality of $A$ there is a $q\in A$ such that $d(p)$ intersects $d(q)$.
If $q\in A_1$ then $d(q)\subset d(p)$ and hence $p\notin R$ because $q\notin R$
and $R$ is downward closed.
In this case we have $f(p)=1$ and $f(q)=0$.
If $q\in A_2$ then take $r\in d(q)\cap d(p)$.
Then $f$ takes on two values on $d(r)$ and hence on $d(p)$.
