Intuition: $\frac{1}{p}+\frac {1}{q} = 1$ equivalent to $(p-1)(q-1)=1$.
It's intuitive if you think of $1/p$ and $1/q$, which we'll denote by $P$ and $Q$.
- The first equation tells you that $P+Q=1$, i.e. you can think of them as probabilities of complementary events, say $P=\operatorname{Prob}(A)$ and $Q=\operatorname{Prob}(\text{not }A)$.
- The second equation, in terms of $P$ and $Q$ is $$\frac{1-P}{P}\cdot\frac{1-Q}{Q}= \frac{\operatorname{Prob}(\text{not }A)}{\operatorname{Prob}(A)} \cdot \frac{\operatorname{Prob}(A)}{\operatorname{Prob}(\text{not }A)}=1$$
I think you can get some pictorial intuition if you rewrite $\frac{1}{p}+\frac{1}{q}=1$ as $q+p=pq$. Pretending $p$ and $q$ are positive numbers greater than $1$ for now, you can draw a rectangle with side lengths $q-1$ and $p-1$ with area $A$, whatever it is. Then extend both legs by $1$ "unit", and make a bigger rectangle. You end up getting two "thinner" rectangles of area $p-1$ and $q-1$ on the sides of your original rectangle, plus a little unit square in the corner. I drew a pretty low quality picture below.
Then geometrically, if $(q-1)(p-1)=1$, i.e., $A=1$, adding the area of four pieces shows the big outer rectangle has area $q+p$, so $pq=q+p$.
Conversely, if $pq=q+p$, so that the big outer rectangle has area $q+p$, the thin upper rectangle and unit square give you $p$ units of area, the side thin rectangle gives you $q-1$ units of area, so $A$ must equal $1$ to give you the last unit of area. That is $(p-1)(q-1)=1$.
