To determine whether there exists a positive integer $k$ such that $p = 6k + 1$ is a prime and
$$\binom{3k}{k} \equiv 1 \pmod{p},$$
we proceed as follows:

Let $g$ be a primitive root modulo $p$. By definition, $g^{6k} \equiv 1 \pmod{p}$. For any integer $a$ such that $p \nmid a$, by Fermat's Little Theorem, we have $a^{6k} \equiv 1 \pmod{p}$. Therefore, $a^{3k} \equiv -1, 0,$ or $1 \pmod{p}$.

Since $g$ is a primitive root modulo $p$, the set $\{ g, g^2, \ldots, g^{6k} \}$ is equivalent to $\{ 1, 2, \ldots, 6k \}$.

Consider the sum
$$A = \sum_{i=1}^k (g^{6i} + 1)^{3k}.$$
This sum can only take values in the set $\{ -k, -k+1, \ldots, k-1, k \} \pmod{p}$.

Expanding $A$ and arranging by binomial coefficients, we get:
$$A = \sum_{j=0}^{3k} \binom{3k}{j} \left[ (g^{6j})^k - 1 + (g^{6k})^{k-2} + \cdots + 1 \right].$$

For $j \neq 0, k, 2k, 3k$, we have $(g^{6j})^k - 1 + (g^{6k})^{k-2} + \cdots + 1 \equiv 0 \pmod{p}$ because $(g^{6j})^k - 1 \equiv 0 \pmod{p}$.

Thus,
$$A \equiv k \left( \binom{3k}{0} + \binom{3k}{k} + \binom{3k}{2k} + \binom{3k}{3k} \right) \pmod{p}.$$

If there exists a prime $p = 6k + 1$ such that $\binom{3k}{k} \equiv 1 \pmod{p}$, then
$$A \equiv 4k \pmod{p}.$$

However, this contradicts the earlier result that $A$ must be in the set $\{ -k, -k+1, \ldots, k-1, k \} \pmod{p}$. Therefore, there is no such prime $p$.

The answer is: \boxed{\text{No, there does not exist such a prime } p.}