If $L$ is a complete lattice and $P$ is a poset and $f: L\to P$ is an order preserving surjective map, does this imply that $P$ is a (complete) lattice?
Not necessarily  there is even a finite counterexample.
Let $X = \{1,2,3\}$ and set $L := \mathcal{P}(X)$.
Let $P := \{b, t\} \cup \{(i,j): i,j \in \{0,1\}\}$, where $b$ will be the botton (least) element, $t$ is the top (greatest) element, and $\{(i,j): i,j \in \{0,1\}\}$ is ordered by $$(i, j) < (k, l) \text{ if and only if } i = 0 \text{ and } k = 1.$$ So $P$ is not a lattice, as $(0,0)$ and $(0,1)$ do not have a least upper bound.
As for the surjection $f: \mathcal{P}(X) \to P$, set $\emptyset \mapsto b$, $X\mapsto t$, and moreover $\{1\} \mapsto (0,0)$ and $\{2\},\{3\}\mapsto (0,1)$ and $X\setminus \{1\} \mapsto (1,0)$ and the rest maps to $(1,1)$.
It's a routine verification that $f$ is onto and orderpreserving.
No, clearly not, because you could put junk on top.
But even if you avoid this by insisting that the map is surjective, there are counterexamples. Consider the map of Eric Wofsey's recent answer, where he considered the partial order $L(X)$, where $X$ is an antichain of pairwise incomparable elements, and we add $0$ and $1$ to bound it. This is a complete lattice, and it admits an order preserving surjection to any bounded partial order on $X$. Such an order may not be a complete lattice, and so this provides numerous counterexamples.

1$\begingroup$ The same shows that $f$ can even be bijective, where it would be the same type of thing as often occurs in topology: a map which is an isomorphism at the level of sets but far from an isomorphism at the level of the structure considered. $\endgroup$ – Todd Trimble♦ Jan 27 '15 at 13:46