Congruences for the second-order Catalan numbers

Let $p$ be any odd prime. We mainly show that

$$\sum_{k=1}^{p-1}\frac{2^{k}}k\binom{3k}k\equiv 0 (\operatorname{mod} p)$$

and

$$\sum _{k=1}^{p-1}2^{k-1} C_{k}^{(2)}\equiv (-1)^{(p-1)/2}-1 (\operatorname{mod} p),$$

where $C_{k}^{(2)}=\binom{3k}k/(2k+1)$ is the $k$th Catalan number of order 2.