Consider an algebraic theory with only one $4$-ary operation $p$, and let $V$ be the variety generated by the equations:
$p(x,x,x,y) \approx x$
$p(x,x,y,x) \approx x$
$p(x,y,x,x) \approx x$
$p(y,x,x,x) \approx x$
does this imply that $V$ is congruence distributive?
Too long for a comment.
"Since, I do not have access to the original journal article, could I ask that the you tell me what the jonsson terms are?"
Jonsson theorem states that $V$ is congruence-distributive if and only if there is a natural number $n$ and there are ternary terms $p_0, p_1, \dots, p_n$ (called Jonsson terms) such that $V$ satisfies the following identities:
\begin{gather}p_0(x, y, z) \approx x, \quad p_n(x, y, z) \approx z,\\ p_i(x, y, x) \approx x \quad (i = 1, 2, \dots, n - 1),\\ p_i(x, x, z) \approx p_{i + 1}(x, x, z) \quad (i \text{ is even and $i \neq n$}),\\ p_i(x, z, z) \approx p_{i + 1}(x, z, z) \quad (i \text{ is odd and $i \neq n$}). \end{gather}
As Eran noted, there is a general result about the varieties with $k$-ary near-unanimity term. I just want to show how to obtain Jonsson terms from the near-unanimity term in your particular case, so you can get the main idea of the proof. Define the following ternary terms: $$p_0(x, y, z) = x, \quad p_1(x, y, z) = p(x, p(x, x, y, z), y, z),\quad p_2(x, y, z) = p(x, p(x, x, z, z), y, z), p_3(x, y, z) = p(x, p(x, y, z, z), y, z), \quad p_4(x, y, z) = z.$$
We have $p_1(x, y, x) = p(x, p(x, x, y, x), y, x) \approx p(x, x, y, x) \approx x$. Same for $p_2, p_3$ and $p_4$. Also $$p_1(x, x, z) = p(x, p(x, x, x, z), x, z) \approx p(x, x, x, z) \approx x = p_0(x, x, z).$$ I leave it up to you to check the remaining identities.
I think the proof is easy, but I forget it. So, I'll just refer you to the original proof:
Mitschke, Aleit Near unanimity identities and congruence distributivity in equational classes. Algebra Universalis 8 (1978), no. 1, 29–32.