Let $d\in\mathbb R^d$ and $p:\mathbb R^d\to(0,\infty)$. Moreover, let $\sigma>0$, $$\tilde p(x):=p(\sigma x)\;\;\;\text{for }x\in\mathbb R^d$$ and $$\hat p:=\frac12\ln\tilde p.$$
Question: Can we find $b\in\mathbb R$ and $c\ge0$ such that $$\phi:=\frac{\|\nabla\hat p\|^2+\Delta\hat p}2-b\in[0,c]?\tag1?$$
If I'm not mistaken, we should have $$\nabla\hat p=\frac12\frac{\nabla\tilde p}{\tilde p}\tag2$$ and $$\Delta\hat p=\frac12\left(\frac{\Delta\tilde p}{\tilde p}-\left\|\frac{\nabla\tilde p}{\tilde p}\right\|^2\right)\tag3.$$ So, $$\|\nabla\hat p\|^2+\Delta\hat p=\frac12\frac{\Delta\tilde p }{\tilde p}-\frac14\left\|\frac{\nabla\tilde p}{\tilde p}\right\|^2\tag4$$ (unfortunately, $\left\|\frac{\nabla\tilde p}{\tilde p}\right\|^2$ did not vanish; did I made a mistake?).
To proceed, I'm willing to assume that $$p_0\le p\le p_\infty\tag4$$ for some $p_0,p_\infty>0$.
As an extension of this question, I would like to know what will happen, when $p$ is not known to be differentiable and we use automatic differentiation instead of analytic differentation for the computation of $\nabla p$ and $\Delta p$.
No, I don't believe this is true unless $p$ also satisfies some PDE which you haven't mentioned. This is saying that a control on the $L^\infty $ norm of $p$ has, in some sense, a control of the $C^2$ norm which is not true in general.
For example, fix $\varepsilon\in (0,1)$ and take $\hat p = \sqrt{x_1^2+\varepsilon}$. The reason I chose this function is because, as $\varepsilon \to 0$, $\hat p \to \vert x_1\vert$ which is not differentiable, so we should expect that $\Delta \hat p $ is unbounded in this limit. By a direct computation, $$ \vert\nabla \hat p(x)\vert^2 = \frac{x_1^2}{x_1^2+\varepsilon} $$ and $$ \Delta \hat p(x) = \frac \varepsilon {(x_1^2+\varepsilon)^{3/2}}. $$ Hence, $$\vert\nabla \hat p(0)\vert^2+ \Delta \hat p(0)= \frac 1 {\sqrt \varepsilon} \to +\infty $$ as $\varepsilon \to 0$. Thus, there cannot exist $b$ and $c>0$ (independent of $\hat p $ and, therefore, $p$) such that $$ \frac 12 \big ( \vert\nabla \hat p(x)\vert^2+ \Delta \hat p(x) \big)-b \leqslant c .$$ You can argue similarly that you cannot have $\vert \nabla \hat p(x)\vert^2+ \Delta \hat p(x) \big)-b \geqslant 0$ by replacing $\hat p$ with $-\hat p$.
Now, this isn't quite a counter-example to your question since, we need to choose $$p(x) = e^{2 \hat p(x/\sigma)} = e^{2 \sqrt{x_1^2/\sigma^2+\varepsilon}}$$ which is unbounded, but this essentially gives the answer. To obtain the true counter-example, we take $\eta \in C^\infty(\mathbb R)$ such that $0\leqslant \eta \leqslant 1$, $\eta=1$ in $(-1,1)$, and $\eta=0$ in $\mathbb R\setminus (-2,2)$ and choose $\hat p(x) = \eta(x_1) \sqrt{x_1^2+\varepsilon}$. The above computations remain valid and now $$ p(x) = e^{2 \eta(x_1/\sigma)\sqrt{x_1^2/\sigma^2+\varepsilon}} \in \big [1,e^{2 \sqrt{4/\sigma^2+1}} \big] $$