Taking the product of the 1st, the 3rd and the 5th fractions reveals that their value has to be 1, that is
$$a + b = c + d = e + f = p. \tag{1}$$
the form of the second and of the fourth fraction implies
$$f + a \mid d + e \quad \text{and} \quad d + e \mid b + c. \tag{2}$$
that is first $f + a$ is at most the arithmetic mean of its multiples,
$$f + a \le \frac{1}{3}((f+a) + (d+e) + (b+c)) = p, \tag{3}$$
and
$$f + a \mid (f + a) + (d + e) + (b + c) = 3p.$$
Thus $f + a$ divides $3p$ and is in the interval $\langle 2, p \rangle$. Consequently either $f + a = p$ or $f + a = 3$. We deal separately with these cases.

i) Let $f + a = p$. Because of (3) there is $f + a = d + e = b + c = p$, which together with (1) gives $p - 1$ solutions of the form
$$(a, b, c, d, e, f) = (a, p-a, a, p-a, a, p-a), \quad \text{where } a \in \{1, 2, \dots, p-1\}.$$

ii) Let $f + a = 3$. Then $\{a, f\} = \{1, 2\}$.

Firstly let $a = 1$ and $f = 2$. According to (1) then $b = p - 1$ and $e = p - 2$, and (2) has the form
$$3 \mid d + (p-2) \quad \text{and} \quad d + (p-2) \mid (p-1) + c. \tag{4}$$
In analyzing (4) we distinguish between $d = 1$ and $d \ge 2$.

If $d = 1$ then $c = p - 1$ and (4) reads
$$3 \mid p-1 \quad \text{a} \quad p-1 \mid 2(p-1).$$
While the right relation always holds, the left one holds only for $p = 3q + 1$ ($q$ is a suitable positive integer). For such prime numbers we get considering (1) solutions
$$(a, b, c, d, e, f) = (1, p-1, p-1, 1, p-2, 2).$$
If $d \ge 2$ we show first, that the right relation in (4) is satisfied if and only if $d + (p-2) = (p-1) + c$ or $d = c + 1$. $d \ge 2$ namely implies $c = p - d \le p - 2$, thus
$$d + (p - 2) \ge p \quad \text{and} \quad (p - 1) + c \le 2p - 3 < 2p$$
and $d + (p - 2) = (p - 1) + c$. From $c + d = p$ and $d = c + 1$ we get $c = \frac{1}{2}(p - 1)$
a $d = \frac{1}{2}(p + 1)$. Since $d + (p - 2) = \frac{3}{2}(p - 1)$, the left relation in (4) is fulfilled and
$$(a, b, c, d, e, f) = (1, p-1, \frac{1}{2}(p-1), \frac{1}{2}(p+1), p-2, 2).$$
is a solution.

Finally $a = 2$ and $f = 1$. In this case $b = p - 2$ a $e = p - 1$, and (2) reads
$$3 \mid d + (p-1) \quad \text{and} \quad d + (p-1) \mid (p-2) + c. \qquad (5)$$
Because
$$d + (p - 1) \ge p \quad \text{and} \quad (p - 2) + c < 2p,$$
the right relation in (5) holds if and only if $d + (p-1) = (p-2) + c$, that is iff $c = d+1$. Together with $c+d=p$ we get $c = \frac{1}{2}(p+1)$ and $d = \frac{1}{2}(p-1)$, thus the right relation in (5) holds as well, and the last solution is
$$(a, b, c, d, e, f) = (2, p-2, \frac{1}{2}(p+1), \frac{1}{2}(p-1), p-1, 1).$$

Conclusion. All the solutions found are apparently mutually different and their number depends on $p$ modulo $3$ ($p > 3$): If $p = 3q + 1$ then there are $p + 2$ sextuples, if $p = 3q + 2$, there are $p + 1$ sextuples.