Solve the congruence system: $p \equiv 11\pmod{24}$ and $ p\equiv 3 \pmod 4$

$$p\equiv11\pmod{24}\implies p\equiv11\pmod4\equiv3$$

So, it is sufficient to have $$p\equiv11\pmod{24}$$

If $p\equiv11\pmod{24}$ then $p\equiv3\pmod{4}$, because if $p=11+24k$ for some $k\in\Bbb{Z}$ then $$p=3+4(2+6k),$$ with $2+6k\in\Bbb{Z}$. So you only need to solve $p\equiv11\pmod{24}$.

The equivalent system, using the Chinese Remainder theorem, should be $$p\equiv2\pmod{3}\qquad\text{ and }\qquad p\equiv3\pmod{8},$$ the latter congruence should be modulo $8$. Breaking down congruences this way is a good approach to solving systems of modular equations, also when the moduli are not coprime.