Whats the number in unit's place?

Relevant wiki: Euler's Theorem

$\begin{aligned} 27^{3^p} - 13^{3^p} & \equiv 7^{\color{#3D99F6} 3^p \text{ mod }\phi(10)} - 3^{\color{#3D99F6} 3^p \text{ mod }\phi(10)} \text{ (mod 10)} & \small \color{#3D99F6} \text{Since }\gcd(7,3,10) = 1 \text{, Euler's theorem applies.} \\ & \equiv 7^{\color{#3D99F6} 3^p \text{ mod }4} - 3^{\color{#3D99F6} 3^p \text{ mod }4} \text{ (mod 10)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(10) = 4 \\ & \equiv 7^{(4-1)^p \text{ mod }4} - 3^{(4-1)^p \text{ mod }4} \text{ (mod 10)} \\ & \equiv 7^{\color{#3D99F6}(-1)^p} - 3^{\color{#3D99F6}(-1)^p} \text{ (mod 10)} & \small \color{#3D99F6} \text{Note that } (-1)^p = \begin{cases} 1 \text{ (mod 4)} & \text{if } p \text{ is even.} \\ -1 \text{ (mod 4)} & \text{if } p \text{ is odd.} \end{cases} \\ & \equiv \begin{cases} 7^1 - 3^1 \equiv 4 \text{ (mod 10)} \\ 7^3 - 3^3 \equiv 7(49)-3(9) \equiv 6 \text{ (mod 10)} \end{cases} \end{aligned}$

Therefore, the answer is either 4 or 6 .