As I mentioned in my comment, when $G$ is connected, you can do this with $B=\mathbb{R}^2$ and $P$ being the trivial bundle $P = G\times\mathbb{R}^2$.

Here is one construction: Let $\frak{g}$ be the Lie algebra of $G$ and let $\gamma:TG\to\frak{g}$ be the canonical left invariant, $\frak{g}$-valued $1$-form on $G$. Let $(r,\theta)$ be polar coordinates on $\mathbb{R}^2$, let $f = f(\theta):S^1\to\frak{g}$ be a curve such that there are values $\theta_1,\ldots,\theta_m$ such that the elements $f(\theta_i)$ for $1\le i\le m$ form a basis for $\frak{g}$, and let $\rho=\rho(r)$ be a smooth, even function of $r$ that vanishes identically in a neighborhood of $r=0$ but has $\rho'(1)=1$. Now consider the connection form $$ \alpha= \gamma + \mathrm{Ad}(g^{-1})\bigl(\rho(r)f(\theta)d\theta\bigr), $$ where $g:P\to G$ is the projection onto the first factor. (When $G$ is a matrix group, this is just $\alpha = g^{-1}dg +g^{-1}\bigl(\rho(r)f(\theta)d\theta\bigr)g$.) Then I claim that the holonomy group of this connection is all of $G$.

Since $G$ is connected, it suffices to show that the Lie algebra of the holonomy group is all of $\frak{g}$ because, by Borel and Lichnerowicz, we know that the holonomy is a Lie subgroup of $G$. To do this, note that the given trivialization is is $\alpha$-parallel along each radial line through the origin $(r=0)$, since along such a line, $d\theta=0$, and $\alpha = g^{-1}dg$. On the other hand, because the curvature is $$ d\alpha + \tfrac12[\alpha,\alpha] = \mathrm{Ad}(g^{-1})\bigl(\rho'(r)f(\theta)\ dr\wedge d\theta\bigr), $$ one sees that, if one takes parallel translation along a curve starting at $\bigl(e,(0,0)\bigr)\in G\times\mathbb{R}^2=P$ that goes out along the radial path from $r=0$ to $r=1$ at angle $\theta_i$, makes a little loop of oriented area $t\approx 0$, around the point $(r,\theta)= (1,\theta_i)$ and then returns to the origin ($r=0$) back along that same radial path, then this will wind up at a point very close to $\bigl(\exp(tf(\theta_i)),(0,0)\bigr)\in G\times\mathbb{R}^2=P$. In particular, $\alpha$-parallel translation of the curvature of $\alpha$ at $(1,\theta_i)$ back to the origin along the radial segment will give you a curvature that takes the value $f(\theta_i)$ at $\bigl(e,(0,0)\bigr)\in G\times\mathbb{R}^2=P$, so the Lie algebra of the holonomy group is all of $\frak{g}$.

More directly, concatenating $m$ of these 'lassoes' about the points $(1,\theta_i)$ with varying values of $t$ will allow you to move freely in every direction at $e\in G$ in the fiber over $(0,0)\in\mathbb{R}^2$. Thus, the holonomy of $\alpha$ will be all of $G$.

As I mentioned in my comment, when $G$ is connected, you can do this with $B=\mathbb{R}^2$ and $P$ being the trivial bundle $P = G\times\mathbb{R}^2$.

Here is one construction: Let $\frak{g}$ be the Lie algebra of $G$ and let $\gamma:TG\to\frak{g}$ be the canonical left invariant, $\frak{g}$-valued $1$-form on $G$. Let $(r,\theta)$ be polar coordinates on $\mathbb{R}^2$, let $f = f(\theta):S^1\to\frak{g}$ be a curve such that there are values $\theta_1,\ldots,\theta_m$ such that the elements $f(\theta_i)$ for $1\le i\le m$ form a basis for $\frak{g}$ and such that $f'$ vanishes in an $\epsilon$-interval about each $\theta_i$ for some $\epsilon > 0$, and let $\rho=\rho(r)$ be a smooth, even function of $r$ that vanishes identically in a neighborhood of $r=0$ but has $\rho'(r)\equiv1$ for $r\ge1$. Now consider the connection form $$ \alpha= \gamma + \mathrm{Ad}(g^{-1})\bigl(\rho(r)f(\theta)d\theta\bigr), $$ where $g:P\to G$ is the projection onto the first factor. (When $G$ is a matrix group, this is just $\alpha = g^{-1}dg +g^{-1}\bigl(\rho(r)f(\theta)d\theta\bigr)g$.) Then I claim that the holonomy group of this connection is all of $G$.

Since $G$ is connected, it suffices to show that the Lie algebra of the holonomy group is all of $\frak{g}$ because, by Borel and Lichnerowicz, we know that the holonomy is a Lie subgroup of $G$. To do this, note that the given trivialization is is $\alpha$-parallel along each radial line through the origin $(r=0)$, since, along such a line, $d\theta=0$, implying that $\alpha = g^{-1}dg$ along such a line. On the other hand, because the curvature is $$ d\alpha + \tfrac12[\alpha,\alpha] = \mathrm{Ad}(g^{-1})\bigl(\rho'(r)f(\theta)\ dr\wedge d\theta\bigr), $$ one sees that, if one takes parallel translation starting at $\bigl(e,(0,0)\bigr)\in G\times\mathbb{R}^2=P$ above the curve that goes out along the radial path from $r=0$ to $r=1$ at angle $\theta_i$, goes counterclockwise around the box $[1,1{+}t]\times[\theta_i,\theta_i{+}\epsilon]$, and then returns to the origin ($r=0$) back along the initial radial path, then this will wind up at the point $\bigl(\exp(tf(\theta_i)),(0,0)\bigr)\in G\times\mathbb{R}^2=P$. (Moreover, $\alpha$-parallel translation of the curvature of $\alpha$ at $(1,\theta_i)$ back to the origin along the radial segment will yield a curvature that takes the value $f(\theta_i)$ at $\bigl(e,(0,0)\bigr)\in G\times\mathbb{R}^2=P$, so the Lie algebra of the holonomy group is all of $\frak{g}$.)

More directly, concatenating $m$ of these 'lassoes' about the points $(1,\theta_i)$ with appropriate values of $t$ will yield an $\alpha$-horizontal curve connecting $\bigl(e,(0,0)\bigr)\in P$ to the point $$ \bigl(\exp(t_1f(\theta_i))\cdots\exp(t_mf(\theta_m)),(0,0)\bigr)\in P. $$ Thus, one can reach an open set in the fiber over $(0,0)$ starting from $\bigl(e,(0,0)\bigr)\in P$ and traveling along $\alpha$-horizontal curves. Thus, since $G$ is connected, the holonomy of $\alpha$ is all of $G$.