Let $p>3$ be an odd prime and $a$ be a positive integer, is the following congruence true?
$$\binom{p^a-1}{\frac{p^a-1}{2}}\equiv(-1)^{\frac{p^a-1}{2}}4^{p^a-1}\pmod{p^3}.$$
When $a=1$, this is Morley's congruence.
Let $p>3$ be an odd prime and $a$ be a positive integer, is the following congruence true?
$$\binom{p^a-1}{\frac{p^a-1}{2}}\equiv(-1)^{\frac{p^a-1}{2}}4^{p^a-1}\pmod{p^3}.$$
When $a=1$, this is Morley's congruence.
The answer is yes. See Corollary 3 of "A congruence involving the quotients of Euler and its applications (I)" by Tianxin Cai (Acta Arithmetica, 2002).
Namely, Cai shows that $$(-1)^{\frac{p-1}{2}} \binom{p^k-1}{\frac{p^k-1}{2}}/\binom{p^{k-1}-1}{\frac{p^{k-1}-1}{2}} \equiv 4^{\phi(p^k)} \bmod {p^{3k}}.$$ In particular, for any $k\ge 1$, $$(-1)^{\frac{p-1}{2}} \binom{p^k-1}{\frac{p^k-1}{2}}/\binom{p^{k-1}-1}{\frac{p^{k-1}-1}{2}} \equiv 4^{\phi(p^k)} \bmod {p^{3}}.$$ By applying the above for all $1 \le k \le a$, he deduces that $$(−1)^{\frac{(p-1)a}{2}} \binom{p^a-1}{\frac{p^a-1}{2}} \equiv 4^{p^a-1} \bmod {p^3}.$$ To see that this is the same as your congruence, note that $(p-1)a \equiv p^a-1 \bmod 4$ by considering two cases: $p \equiv 1 \bmod 4$ and $p \equiv -1 \bmod 4$.