One way to approach this problem could be to imagine zero as a very small number.
So to start with, let us define,
$$small \to 0$$
$$\therefore$$
instead of $$ (2 \cdot 0) \cdot p = 1 $$
as above @Jim H. Use:
$$2 \cdot small \cdot \frac{1}{small} = 2$$
As
$$(2 \cdot small) \cdot \frac{1}{small} \neq small \cdot \frac{1}{small} = 1$$
$$= 2 \cdot \frac{small}{small} \cdot 1 = 2$$
Do not use $(2 \cdot small) \to 0$ during multiplication and division, otherwise it will produce nonsense such as 2 = 1.
$\underline {The\, p\, number\, system}$
If a very small number, $$s \to 0$$ $$p = \frac{1}{s} \to \infty$$
then, $$\frac{p}{p} = \frac{1}{s} \cdot \frac{s}{1} = 1$$
$$\therefore p \cdot p = \frac{1}{s} \cdot \frac{1}{s} = \frac{1}{s^2} = p^2$$
Also, we could invent a p number,
$$4 + 3p$$
multiply it by s,
$$(4 + 3p) \cdot s = 4 \cdot s + 3 \cdot \frac{1}{s} \cdot s$$
$$4 \cdot s \to 0$$
$$\therefore (4 + 3p) \cdot s = 3$$
$\underline {Adding\, and\, multiplying\, fractions\, in\, p\, number\,system}$
$$p + p = \frac{1}{s} + \frac{1}{s} = \frac{1 + 1}{s} = 2p$$
$$p - p = \frac{1}{s} - \frac{1}{s} = \frac{1 - 1}{s} = \frac{0}{s} = 0$$
$$\frac{2}{3} + \frac{p}{3} = \frac{2 + p}{3}$$
Instead of $$\frac{0}{0} = 0$$ use $$\frac{s}{s} = 1$$
$$\frac{2}{3} + \frac{s}{s} = \frac{2}{3} + 1 = \frac{5}{3}$$
$\underline {Adding\, two\, p\, numbers}$
$$(3 + 2p) + (5 + p) = 8 + 3p$$