Fourier Mukai transform of skyscraper sheaf through flat kernel. Let $X,Y $ be smooth projective varieties over a field $k$.
The following example is taken from Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" [example 5.4.vi]:
"Suppose $\mathcal{P}$ is a coherent sheaf on $X\times Y$ flat over $X$ and consider the Fourier Mukai Transform $\Phi_\mathcal{P}$. If $x\in X$ is a closed point with $k(x)\cong k$, then $\Phi_\mathcal{P}(k(x))\cong P_{|x\times Y}$ where $\mathcal{P}_{|x\times Y}$ is considered as a sheaf on $Y$ via the second projection."
Let $q: X\times Y\to X$ and $p: X\times Y\to Y$ denote the projections. Let $i: x\times Y\to X\times Y$ denote the closed immersion obtained from $x\to X$ by base change.
I calculated $q^*k(x)\cong i_*i^*\mathcal{O}_{X\times Y}$ and then I got 
$q^*k(x)\otimes\mathcal{P}\cong i_*i^*\mathcal{O}_{X\times Y}\otimes \mathcal{P}\cong i_*(i^*\mathcal{O}_{X\times Y}\otimes i^*\mathcal{P})\cong i_*i^*\mathcal{P}$ via the projection formula.
Applying the right derived functor $Rp_*$ yields $Rp_*i_*i^*\mathcal{P}\cong R(p_*i_*)(i^*\mathcal{P})\cong q'_*i^*\mathcal{P}$ where $q':x\times Y\to Y$ is the second projection. This is the result from the example.  I do not see where I have used flatness of $\mathcal{P}$ over $X$ though. Where do I need it ? Thanks a lot .
EDIT: Flatness over $X$ should play a role in calculating the derived tensor. I actually assumed the derived tensor product equals the usual tensor product when calculating the FMT. 
Is this implied by $\mathcal{P}$ being flat over $X$?
 A: I know it’s a very old question, but I think I figured out what’s the solution, so I post the answer.
The final answer you get is $q’_* i^* \mathcal{P} $, however, you don’t know whether this is a sheaf or a true complex. First we notice that $q’_*$ need not be derived being an isomorphism, as the closed point was chosen such that $k(x) \simeq k$ (I’m quite sure about this, as it works affine locally, but correct me if I’m wrong). As $\mathcal{P}$ is flat over $X$ we know that the functor $\mathcal{G} \mapsto q^*\mathcal{G} \otimes \mathcal{P}$ from modules over $X$ to modules over $X \times Y$ is exact, where $q$ is the projection. Therefore, $q^* k(x) \otimes \mathcal{P}$ is a sheaf and the tensor product is not derived. Hence, we can use the stand projection formula (not the derived one) as $i_*$ is also exact, and therefore not derived, to get 
$$
q^* k(x) \otimes \mathcal{P} \simeq i_* i^* \mathcal{P}
$$
as sheaves. We now conclude applying the functor $Rp_*$, which yields
$$
\Phi_{\mathcal{P}} (k(x)) \simeq q’_* (i^* \mathcal{P})
$$
as a sheaf.
