The answer is no. Here is a counterexample that Agnes Szendrei and I found.
Consider the reflexive, symmetric, binary relation $T = \{0,m\}^2\cup \{m,1\}^2$ on the set $S=\{0,m,1\}$. The counterexample is the algebra $A$ whose universe is $S$ and whose operations are all operations that preserve $T$. (I.e., all operations $f$ on $S$ where $T$ is a subalgebra of $\langle S,f\rangle^2$.) This algebra is clearly not primal, since $T$ is a compatible binary relation of $A$ that is different from the total binary relation or the equality relation, and a primal algebra has no such compatible relation.
The lattice operations for the order $0<m<1$ preserve $T$, so $A$ is an expansion of the 3-element lattice. It is easy to check that $A$ is finite, simple, has no proper subalgebra, has no automorphism except the identity map, and has constant operations naming each of its elements. It follows that the variety $\mathcal V$ generated by $A$ is semisimple and congruence distributive. It then follows from Theorems 2.12 and 2.21 of
Ervin Fried and Emil W. Kiss, Connections between congruence-lattices and
polynomial properties, Algebra Universalis, 17 (1983) 227-262.
that $\mathcal V$ has EDPC.
What remains is to prove that $A$ is $c$-regular for $c = 0, m, 1$.
I argue by contradiction. I only explain $0$-regularity, but the other two cases are similar. Assume that $B\in \mathcal V$ has congruences $\alpha\neq\beta$ such that $0/\alpha=0/\beta$. Replacing one of the congruences with their intersection and renaming we may assume that $\alpha < \beta$. Factoring by $\alpha$ we may assume it is zero. Now we have a congruence $\beta\neq 0$ with $0/\beta = \{0\}$. The properties that hold for $B$ and $\beta$ can be realized in a finite subalgebra of $B$, so assume that $B$ is finite. Now we can assume that $|B|$ is minimal and that $\beta$ is minimal in $Con(B)$ among algebras in $\mathcal V$ realizing this data. To summarize and refine: we may assume that $B\leq A^n$ is an irredundant subdirect power of $A$ and that $\beta$ is an atom in $Con(B)$. The atoms are the restrictions of the kernels of the projections onto all but one of the coordinates, so we may assume that $\beta$ relates tuples that are equal in all but the first coordinate.
Now we use the fact that $A$ has a majority operation (from the lattice structure) and that $T$ is the only proper nonequality compatible relation of $A$ (check). This tells us what the possible finite, irredundant subdirect powers of $A$ are. The algebra $B\leq A^n$ is definable by a simple graph (say $G$) on the index set $\{0,1,\ldots,n-1\}$. Each edge $\{i,j\}$ of $G$ imposes a restriction on $ij$-th projection of $B$: the pair $(b_i,b_j)$ must lie in $T$. Thus, $B$ is the set of all tuples $(b_0,\ldots,b_{n-1})$ such that $(b_i,b_j)\in T$ whenever $\{i,j\}$ is an edge in $G$. (If what I have written sounds confusing, let me summarize the key point as follows: the Baker Pixley Theorem implies that any irredundant subdirect subalgebra of $A^n$ contains $\{0,m\}^n\cup \{m,1\}^n$.)
Now it is easy to see from the above description that the tuple $t=(m,0,0,\ldots,0)$ belongs to $B$ no matter what $B$ is. Moreover, this tuple differs from $0=(0,0,0,\ldots,0)$ in the first coordinate only. I.e., the $\beta$-class of $0$ is not $\{0\}$ after all, since it contains $t\neq 0$.
