The answer is no.
Take the "classical" example of the line with two origins. This space is non-Hausdorff, paracompact and doesn't admit partitions of unity.
EDTI: I think the question is a kind of "duplicate" .
Ok, but if you have an example for a non-Hausdorff manifold, which doesn't admit partitions of unity, you have an example for a non-Hausdorff paracompact space with the same property.
First the definition:
The line with two origins is the quotient space of two copies of the real line
$mathbb{R} times {a}$ and $mathbb{R} times {b}$.
with equivalence relation given by
$(x,a) sim (x,b)text{ if }x neq 0$.
Since all neighbourhoods of $0_a$ intersect all neighbourhoods of $0_b$, it is non-Hausdorff.
However, this space is paracompact, since $mathbb{R}$ is paracompact.
For the non-existence of a partition of unitiy: take the open covering $ U = (-infty,0) cup { 0_a } cup (0,infty)$ and $tilde{U} = (-infty,0) cup { 0_b } cup (0,infty)$. Assume, there is a partition of unity subordinate to this cover. Then the value of each origin would have to be $1$ which cannot be true. (Edit: villemoes was a little faster :-) )
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