Supercongruences involving products of two binomial coefficients

In this paper we deduce some new supercongruences modulo powers of a prime $p>3$. Let $d\in \{0,1,\dots ,(p-1)\mathrm{/}2\}$. We show that $$\sum _{k=0}^{(p-1)\mathrm{/}2}\frac{\left(\genfrac{}{}{0px}{}{2k}{k}\right)\left(\genfrac{}{}{0px}{}{2k}{k+d}\right)}{8^k}\equiv 0(modp)ifd\equiv \frac{p+1}{2}(mod2),$$ and $$\sum _{k=0}^{(p-1)\mathrm{/}2}\frac{\left(\genfrac{}{}{0px}{}{2k}{k}\right)\left(\genfrac{}{}{0px}{}{2k}{k+d}\right)}{{16}^k}\equiv \left(\frac{-1}{p}\right)+p^2\frac{(-1{)}^d}{4}{E}_{p-3}(d+\frac{1}{2})(\mathrm{m}\mathrm{o}\mathrm{d}{p}^3),$$ where $E_{p-3}(x)$ denotes the Euler polynomial of degree $p-3$, and $(-)$ stands for the Legendre symbol. The paper also contains some other results such as $$\sum _{k=0}^{p-1}{k}^{(1+(\frac{-1}{p}))\mathrm{/}2}\frac{\left(\genfrac{}{}{0px}{}{6k}{3k}\right)\left(\genfrac{}{}{0px}{}{3k}{k}\right)}{{864}^k}\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{p}^2)\mathrm{.}$$