The abstract kernel theorem implies Schwartz kernel theorem. How exactly?

Let me first give a little rapid background prior to formulating the question.
Let $\mathcal{D}$ be a Schwartz space of infinitely differentiable functions and $\mathcal{D}'$ is the space of distributions acting on $\mathcal{D}$.
The Schwartz Kernel Theorem states that for any linear continuous operator $A\colon \mathcal{D}\to \mathcal{D}'$ there exists a distribution $K\in [\mathcal{D}\times \mathcal{D}]'$ such that
$$(Af,\varphi)=K(f\otimes \varphi),$$
where $f\otimes g$ is a tensor product of functions $f$ and $g$.
Grotendieck in “Produits tensoriels topologiques et espaces nucléaires” , Amer. Math. Soc. (1955) obtained a following generalization:
Let $F$ be a nuclear locally convex space. Then for every $E$ locally convex space all continuous bilinear form $B(f,\varphi)$ are nuclear.
A bilinear form $B(f,\varphi)$ on the Cartesian product $F\times E$ two locally convex spaces is called nuclear if it can be represented in the form
$$B(f,\varphi) = \sum\limits^{\infty}_{k=1}\lambda_k(f,f_k)(\varphi,\varphi_k),$$
where $f_k$ and $\varphi_k$ are the sequences in the dual space $\mathcal{D}'$ and $\lambda_k$ is a summable sequence.
Now the question is how to see that if an operator $A$ is continuous between $\mathcal{D}$ and $\mathcal{D}'$,i.e. $A\in\mathcal{L}(\mathcal{D},\mathcal{D}')$, then the bilinear form $B(f,\varphi)=(Af,\varphi)$ is nuclear?

Answer

Attribution
Source : Link , Question Author : Rauan Akylzhanov , Answer Author : Community