Suppose $G$ is a finitely-generated group and $A$ is a finite symmetric set of its generators. Define $\pi: A^* \to G$ using the following recurrence:
$$\pi(\Lambda) = e$$ $$\pi(a \alpha) = a\pi(\alpha)$$
Then we have two computational problems:
Word problem:
Given a word $a \in A^*$ determine whether $\pi(a) = e$ ($e$ is the group identity element).
Conjugacy problem:
Given two word $a, b \in A^*$ determine whether $\pi(a)$ and $pi(b)$ are conjugate.
There are known quite many groups $G$, for which the word problem can be solved in polynomial time (in respect to the length of the word). This class of groups includes all hyperbolic groups, automorphism groups of all free groups, all automatic groups and also many groups, that do not fall under any of these categories (like, for example, the Baumslag-Gersten group $\langle a, b | a^{a^b}= a^2 \rangle$).
For all such groups, the conjugacy problem belongs to NP. Indeed, if we take a word $c \in A^*$ such that $\pi(c)^{-1}\pi(a)\pi(c) = \pi(b)$ as a witness, we can verify it in polynomial time by solving the word problem for $c^{-1}acb^{-1}$.
However, does there exist a group, in which the word problem is polynomial and the conjugacy problem is NP-complete?
Such group (if it exists) will be neither hyperbolic nor automatic (assuming $P \neq NP$), because for all groups from those classes the conjugacy problem is also solvable in polynomial time.
A quick Google search gives this paper and this slightly earlier paper, which constructs groups with NP-complete conjugacy problem and polynomial time word problem.
(The second has cubic word problem since polycyclic groups are linear, so the word problem is reduced to a matrix computation.)