There are a lot of correct answers here, but I think that there is a fundamental definition or intuition that is missing from all of them, namely that we should ignore the value of the expression at the limit point (i.e. we assume that $p$ is never actually zero; we are taking a limit as $p$ approaches zero). A good definition of a limit is as follows:
Definition: We say that $\lim_{x\to a} f(x) = L$ if for all $\varepsilon > 0$ there exists some $\delta > 0$ such that if $x\ne a$ and $|x-a| < \delta$, then $|f(x) - L| < \varepsilon$.
Topologically (and feel free to ignore this paragraph for now), we are saying that for any neighborhood $V$ of $L$, there is some punctured neighborhood $U^\ast$ of $a$ such that $f(U^*) \subseteq V$. Because we are puncturing the neighborhood, the value of $f$ at $a$ is irrelevant. We just completely ignore it.
In the original question, we are trying to compute
$$ \lim_{p\to 0} \frac{1-p-(1-p)^3}{1-(1-p)^3}. $$
As you have noted, when $p=0$, this expression is utter nonsense. That is, if we define
$$ f(p)
:= \frac{1-p-(1-p)^3}{1-(1-p)^3}
= \frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p} $$
then try to evaluate $f(0)$, this will give us $\frac{0}{0}$ which is a (more-or-less) meaningless expression. However, we are trying to take a limit as $p\to 0$, which means that we can (and should) assume that $p \ne 0$. Notice that under this assumption, i.e. the assumption that $p\ne 0$, we have that $1 = \frac{1/p}{1/p}$. Then, using the analyst's second favorite trick of multiplying by 1 (adding 0 is the favorite trick), we have
\begin{align}
f(p)
&= \frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p} \\
&= \color{red}{1} \cdot \frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p} \\
&= \color{red}{\frac{1/p}{1/p}} \cdot \frac{p^3 - 3p^2 + 2p}{p^3 - 3p^2 + 3p} \\
&= \frac{p^2-3p+2}{p-3p+3} \\
&=: \tilde{f}(p).
\end{align}
Again, the vital thing to understand is that the computation is justified since $p \ne 0$, which means that the fraction $\frac{1/p}{1/p}$ is perfectly well-defined and is (in fact) identically 1. Note, also, that the computations above are done before we've tried to take any limits.
It is now relatively easy to see that
$$ \lim_{p\to 0} f(p)
= \lim_{p\to 0} \tilde{f}(p)
= \lim_{p\to 0} \frac{p^2-3p+2}{p-3p+3}
= \frac{2}{3}. $$
There are two things here that I have left unjustified:
Exercises:
- Explain why $\lim_{p\to 0} f(p) = \lim_{p\to 0} \tilde{f}(p)$?
- Explain why $\lim_{p\to 0} \tilde{f}(p) = \frac{2}{3}$.
Hint for 1:
One possible argument is a one-line appeal to the squeeze theorm.
Hint for 2:
Think about the relation between continuity and limits.