For q prime, enlarge $\{ 1,\cdots,m \}$ to a set of size $n=m+(q-1)$ by replacing $m$ by $q$ clones $m_1 , m_2 , \cdots , m_q$ and consider the $q$-cycle $\sigma=(m_1\ m_2\ \cdots \ m_q)$. It acts on the set of partial orders of the $n$-set and each of its orbits has size 1 or size q. Each orbit of size 1 arises from a unique partial order of the $m$-set by having all $p$ clones behave identically to the original. This proves that $p(m+(q-1)) \equiv p(m) \mod q $ I think I see how to generalize to $q^k$ but I'll have to think about it. The same idea should apply to a wider variety of structures, but which ones?
later The argument seems as if it should work for bipartite graphs on n labelled vertices and also connected bipartite graphs except for powers of 2 The data at OEIS supports this as far as it goes, ignoring the numbers for less than 3 vertices. http://www.research.att.com/~njas/sequences/A047864 http://www.research.att.com/~njas/sequences/A001832
I would think it would work
It also works for appropriate restricted classes such as series parallel networks with n labelled edges vertices and parallel edges allowed. Maybe I don't understand the definition because the data seems to show that $m+(q-1)$ edges allows twice as many networks ($\mod q$) as $m$ edges. http://www.research.att.com/~njas/sequences/A006351http://www.research.att.com/~njas/sequences/A053554
Here is my argument for why $p(n+\phi(q^2)) \equiv p(n) \mod q^2$. I think it generalizes to $q^k$: Further enlarge the $n$ set above to one of size $m+q^2-1=n+\phi(q^2)=N$ by replacing each clone $m_i$ by $q$ clones $m_{i1}, m_{i2}, \cdots ,m_{iq}$ and consider the $q^2$ cycle $$\tau=(m_{11}m_{21}\cdots m_{q1}m_{12}m_{22}\cdots m_{q,q})$$ It acts on partial orders of the $N$-set and the action has orbits of size 1, $q$ and $q^2$. The orbits of size less than $q^2$ are in bijective correspondence with the orbits of the same size for the action of $\sigma$ on partial orders of the $n$-set.

