earthmassnewtonian-gravityweight
I was wondering how mass and weight are different so I Googled it. I found that mass is constant no matter where you are because it is the amount of matter in an object and weight changes because it's the pull of gravity on an object.
This lead me to another question. If you are on Earth are your mass and weight the same?
There's a thing called a "slug". " It is a mass that accelerates by 1 ft/s2 when a force of one pound-force (lbF) is exerted on it." (wikipedia).
Sometimes you'll see reference to a "pound-mass" to indicate a mass which weighs one pound at sea level (on Earth, thank you! :-) ).
The problem is that pounds and kilograms have been used colloquially since forever to describe the weight of objects. Scientific usage differs from informal usage such as "shipping weight".
As you rightly pointed out, the fact that 67P is oddly-shaped should alter its gravitational attraction on various parts of the comet.
That said, if we were to go by Wikipedia in a rather off-hand manner, we find that the lander is $100 kg$ (as @fibonatic rightly pointed out) and 67P has an acceleration due to gravity of $\textbf{g'} = 10^{-3} m/s^2$. Its weight $W$ would therefore be simply a calculation of $W = m\,\textbf{g'}$, giving us $W = \frac{10^2}{10^3} kg = 0.1 kg$ or $100 \,\,\verb+earth+\, g$.
[P.S. I will update this answer with better sources than Wikipedia as soon as I find time.]
Edit 1: This ESA webpage seems to confirm the figures.
Edit 2: Calculating $\textbf{g'}$
I made some calculations: Using the formula $\textbf{F} = M\,\textbf{g'} = \frac{GmM}{r^2} \Rightarrow \textbf{g'} = \frac{GM}{r^2}$ we can calculate the acceleration due to gravity on 67P (m being the comet's mass and M that of our lander). The above ESA page gives us this figure:
Seeing how the dimensions vary wildly, I decided to consider a mean of, say, 3.5km as diameter and 1.75km as $r$. 67P's mass is, of course, $10^{13}\,kg$ which gives us, $$ \textbf{g'} = \frac{6.67 \times 10^{-11} \times 10^{13}}{\left( 1.75 \times 10^3 \right)^2} \approx 10^{-3} ms^{-2} $$ A more precise answer, is, of course, $0.217 \times 10^{-3} ms^{-2}$ but since we have been very liberal in our assumptions of mass and radius, I think we ought to simply consider the order of magnitude, $10^{-3} ms^{-2}$. This pdf file contains some simulation data that agrees with our result.
Best Answer
From a unit analysis point of view mass and weight are different kinds of things: mass is a quantity of material substance and weight is a force. Things of a different kind can never be the same.
That said, historically the distinction was not always recognized and traditional systems of weight-and-measure (including the original metric system, but not SI) use the same unit for both.
In those systems of measurement, there have the same numeric value on Earth by construction.