Given that $\xi = \{\text{integers } x : p \le x < 20\}$ $A = \{\text{factors of 24}\}$ $B = \{\text{prime numbers}\}$ $C = \{\text{perfect squares}\}$

1. List the factors of 24 and keep only those in the universal set when $p=2$.

When $p=2$, $\xi=\{\text{integers }x:2\le x<20\}$.
The positive factors of $24$ are $1,2,3,4,6,8,12,24$.
Keeping only those in $\xi$, exclude $1$ and $24$.
Hence
$A=\{2,3,4,6,8,12\}$.

1. Identify the integers below 20 which are neither prime nor perfect squares, starting from $p\ge0$.

For integers $x$ with $0\le x<20$:

Primes: $2,3,5,7,11,13,17,19$

Perfect squares: $0,1,4,9,16$

So the integers that are neither prime nor perfect squares are
$6,8,10,12,14,15,18$.

2. Choose $p$ so that the complement has exactly 4 elements and list them.

Since xi=\{x:p\le x<20\}, increasing $p$ removes smaller integers from consideration.

To have exactly 4 elements in $(B\cup C)'$, keep the last 4 numbers from the list:
$12,14,15,18$ would occur if $p>10$.

But if $p=12$, then
$(B\cup C)'=\{12,14,15,18\}$, so $n((B\cup C)')=4$.

1. Find which factors of 24 are also perfect squares.

Factors of $24$ that are perfect squares are $1$ and $4$.

So, before applying the universal set restriction,
$A\cap C=\{1,4\}.$

2. Choose the smallest lower bound $p$ that excludes all common elements from the universal set.

The universal set is
$\xi=\{\text{integers }x:p\le x<20\}.$

For $A\cap C=\emptyset$, both $1$ and $4$ must be excluded from $\xi$.

The smallest integer lower bound that excludes $4$ is
$p=5.$