A hypergeometric supercongruence for even d

Let $(x)_k$ denote the rising factorial, with $(x)_0=1$. Let $d\geqslant 2$ be an even integer, and let $p\equiv d+1\pmod{2d}$ be a prime. The hypergeometric supercongruence.

$$\sum_{k=0}^{p-1} (3dk+1)\frac{(\frac{1}{d})_{k}^{2}(\frac{d-1}{d})_{k}(\frac{1}{2d})_{k}} {k!^{3}4^{k} (\frac{d+2}{2d})_{k}} \equiv 0 \pmod {p^3}.$$

This is proposed as a variation of an earlier supercongruence obtained by extending the summation range to $p-1$; its validity is supported by numerical calculation, but no proof is supplied here.