A five-digit number from $X$ is a multiple of $5$ iff its last digit is $0$ (since $5$ is not available). With $0$ fixed in the units place, the first four positions are filled by some four-element multiset drawn from $\{1,2,2,2,4,4\}$.

Counting the arrangements for each multiset choice:

• $\{1,2,2,2\}$: $\frac{4!}{3!}=4$
• $\{1,2,2,4\}$: $\frac{4!}{2!}=12$
• $\{2,2,2,4\}$: $\frac{4!}{3!}=4$
• $\{2,2,4,4\}$: $\frac{4!}{2!2!}=6$
• $\{1,2,4,4\}$: $\frac{4!}{2!}=12$

Total multiples of $5$: $4+12+4+6+12=38$.

Multiples of $20$ end in $20$ or $40$ (since $00,60,80$ are unavailable).

End in $20$: remove one $2$ and the $0$; pick three from $\{1,2,2,4,4\}$ for the first three positions. Arrangement counts: $\{1,2,2\}\to 3,\{1,2,4\}\to 6,\{1,4,4\}\to 3,\{2,2,4\}\to 3,\{2,4,4\}\to 3$, total $18$.

End in $40$: remove one $4$ and the $0$; pick three from $\{1,2,2,2,4\}$. Counts: $\{1,2,2\}\to 3,\{1,2,4\}\to 6,\{2,2,2\}\to 1,\{2,2,4\}\to 3$, total $13$.

Multiples of $20$: $18+13=31$. Hence $p=\frac{31}{38}$ and $38p=31.$