Can the cofinality of a $\operatorname{cf}(A)$-directed partially ordered set $A$ be singular? The cofinality of a totally ordered set is always a regular cardinal. On the other hand for any cardinal (regular or singular) $\kappa$ there is a partially ordered set $(A,\leq)$ with $\operatorname{cf}(A,\leq)=\kappa$, for example $(\kappa,=)$.
Assume that $(A,\leq)$ is a partially ordered set of infinite cofinality $\kappa$, and furthermore that $(A,<)$ is $\kappa$-directed, meaning that if $B\subset A$ and $|B|<\kappa$ then there is $a\in A$ such that $b<a$ for all $b\in B$. With this additional assumption, is it still possible that $\kappa$ is a singular cardinal?
Added question: If we change the additional assumption of $\kappa$-directedness to the assumption that $(A,\leq)$ is a directed set, is it even then possible that $\operatorname{cf}(A,\leq)$ is a singular cardinal? In other words, is there a directed partially ordered set whose cofinality is a singular cardinal?
 A: Here’s a positive answer to the added question.
Let $\kappa$ be an infinite cardinal. Let $D_0 = \kappa$. Given $D_n$ for some $n \in \omega$, for each $p \in [D_n]^2$ let $E_{n+1} = [D_n]^2 \times \kappa$, $D_{n+1} = D_n \cup E_{n+1}$, and $R_n = \{\langle a,\langle \{a,b\},\alpha \rangle\rangle:\{a,b\} \in [D_n]^2 \land \alpha \in \kappa\}$. Let $D = \bigcup\limits_{n\in\omega} D_n$ and $R = \bigcup\limits_{n\in\omega} R_n$, and let $\preceq$ be the transitive closure of $R$. Clearly $\langle D,\preceq \rangle$ is a directed set, and $|D| = \kappa$. For each $d \in D$ let $\lceil d \rceil = \{a \in D:a \preceq d\}$; clearly $\lceil d \rceil$ is finite. 
Now suppose that $A \subseteq D$ with $\omega \le |A|<\kappa$. Let $\lceil A \rceil = \bigcup\limits_{a\in A} \lceil a \rceil$; then $|\lceil A \rceil| = |A| < \kappa = |D|$, so $A$ cannot be cofinal in $D$, and hence $\operatorname{cf}D = \kappa$. This is true even for singular $\kappa$.
Added: As LostInMath suggests, this is working too hard: $E_{n+1}$ need only contain one upper bound for each $p \in [D_n]^2$, so we could simply let $E_{n+1} = [D_n]^2$, or avoid $E_{n+1}$ altogether and just let $D_{n+1} = D_n \cup [D_n]^2$.
