I think I can provide some intuition for (A), both in characteristic $0$ and $p > 0$. What follows below is more or less a proof, but with a lot of omissions (and hopefully not too many lies...).
I believe that everything I state works for all (geometrically) integral projective $k$-schemes. For simplicity, let's assume $k = \bar k$.
Reference. A great reference for the Grothendieck-style approach to this theory is FGA Explained, chapter 9, or Néron Models, chapter 8. (For general discussion of deformation theory and Hilbert and Quot schemes, I would recommend FGA Explained, whereas for the actual construction of the Picard scheme maybe Néron Models is a bit more readable.)
For a $k$-scheme $X$, we can view the tangent space at a closed point $x \to X$ as the extensions
$$x \to \operatorname{Spec} k[\varepsilon]/\varepsilon^2 \to X.$$
That is, it is the preimage of $x \in X(k)$ under the map $\phi\colon X(k[\varepsilon]/\varepsilon^2) \to X(k)$. If $X$ is a group scheme, then $\phi$ is a group homomorphism, and the tangent space at the identity corresponds to the kernel of $\phi$.
Now apply this to the Picard group scheme $\operatorname{\underline{Pic}}_X$. This is usually defined as the fppf sheafification of the functor $Y \mapsto \operatorname{Pic}(X\times Y)/\operatorname{Pic}(Y)$, but in the projective case it suffices to take the étale sheafification; the big difficult theorem is that this is representable by a scheme. (I believe that the sheafification takes care of dividing out by $\operatorname{Pic}(Y)$, so you could omit that step if you want to.)
Thus, we have to compute line bundles on $X_\varepsilon := X \times \operatorname{Spec} k[\varepsilon]/\varepsilon^2$ that are trivial on the central fibre $X$. (Since $k[\varepsilon]/\varepsilon^2$ has no étale covers, we don't need to worry about the sheafification business.) On $X_\varepsilon$, we have a short exact sequence
$$0 \to \mathcal O_X \to \mathcal O_{X_\varepsilon} \to \mathcal O_X \to 0,$$
coming from $0 \to k \to k[\varepsilon]/\varepsilon^2 \to k \to 0$. Tensoring with a line bundle $\mathcal L$ trivial on $X$ gives
$$0 \to \mathcal O_X \to \mathcal L \to \mathcal O_X \to 0.$$
Conversely, any such extension can be given the structure of line bundle on $X_\varepsilon$ by having $\varepsilon$ act by the composition $\mathcal L \to \mathcal O_X \to \mathcal L$. Hence, we are classifying extensions of $\mathcal O_X$ by itself. These are given by
$$\operatorname{Ext}^1_X(\mathcal O_X,\mathcal O_X) = H^1(X,\mathcal O_X).$$
Thus, we have proven:
Theorem. The tangent space $T_0 \operatorname{\underline{Pic}}_X$ equals $H^1(X,\mathcal O_X)$. $\square$
In characteristic $0$, any finite type group scheme is smooth, so we get the equality of dimensions. In characteristic $p > 0$, there are non-smooth group schemes, the simplest example being
$$\mu_p = \operatorname{Spec} k[x]/(x^p-1) \subseteq \mathbb G_m.$$
In those cases, you actually get the wrong dimension:
$$\dim H^1(X,\mathcal O_X) \geq \dim \operatorname{\underline{Pic}}_X,$$
with strict inequality if $\operatorname{\underline{Pic}}_X$ is singular.
