Entropy paradox and Confusion with Gibbs energy
The essence of what's going on is this:
Your equation, $dG = Vdp -SdT$, is correct, but has restrictions you don't realize. Specifically, it applies only to systems that are (a) closed (so no adding or subtracting material), (b) have only a single component* (so no mixing, phase changes, or chemical reactions) and (c) that can do $pV\ \hspace{-.45em}$-work only.
[*It also applies to multi-component systems where the composition is fixed.]
To understand why, under such conditions, $dT = 0 \text{ and } dp = 0 \Rightarrow dG = 0$, let's apply the Gibbs phase rule to such a system.  The phase rule says:
$$F = C − P + 2,$$
where C is the number of components, P is the number of phases at equilibrium (not to be confused with $p$, pressure), and F is the number of degrees of freedom.
Since $C = 1 \text{ and } P = 1$, we obtain $F = 2$.  That means we only have two degrees of freedom, i.e., two independent ways in which we can vary the intensive properites of the system. If the only type of work we can do is $pV\ \hspace{-.45em}$-work, the only ways we can adjust the system's intensive properties are by changing its temperature or its pressure.
Hence, if we apply the restrictions $\boldsymbol{dT=0 \textbf{ and } dp =0}$ to a closed single-component system where we only allow $\boldsymbol{pV}$-work, the system can't change!  And if the system can't change, then of course $\boldsymbol{dG = 0}$!
But, you might protest, $dG$ is not generally zero, even at fixed  $T\text{ and } p$.  So how do we reconcile this with what we wrote above?  Well, we need a more general expression for $dG$ that allows for non-$pV\ \hspace{-.45em}$-work, addition and subtraction of material, and changes in composition:
$$dU = \text{đ}q + \text{đ}w +\sum_i \mu_i dn_i 
= \text{đ}q + \text{đ}w(pV) + \text{đ}w (non\text{-}pV) + \sum_i \mu_i dn_i$$
Since we can calculate $dU$ using any path, let's use a reversible path:
$$dU = TdS - pdV + \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
And since:
$$G = U+pV-TS \Rightarrow dG = dU + pdV +Vdp - TdS - SdT$$
$$\Rightarrow  dG =  VdP -SdT+ \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
Here, $\sum_i \mu_i dn_i$ is the sum of the chemical potential of each species $i$  $(\mu_i)$ times the change in the amount of species $i$ $(dn_i)$. This accounts for the change in $U$, and thus in $G$, as we change the composition
Hence, even if $dT = 0 \text{ and } dp = 0$, if we have non-$pV\ \hspace{-.45em}$-work and/or a change in composition (e.g., mixing, a change in phase, gain or loss of material, or a chemical reaction), it will not be the case that $dG$ is restricted to be zero.