Yes, if $X$ is smooth and $\bar X\cap \mathbb P^{n-1}$ is smooth in the scheme theoretic sense, then $\bar X$ is indeed smooth.
In other words if $s\in \bar X$ is singular, so is $s\in \bar X\cap H $ for any hyperplane $s\in H\subset \mathbb P^{n}$ .
Beware however that you have to take scheme-theoretic intersections.
For example if $\bar X$ is the curve $z^2y=x^3$ in $\mathbb P^{2} $, then that curve is smooth in the affine plane $z\neq1$ and its intersection with the line at infinity $H$ given by $z=0$ is the single point $s=[0:1:0]$.
So, naïvely one might think that since a single point is a smooth variety one may conclude by the above that $\bar X$ is smooth.
In reality $\bar X$ has a singularity at $s$. The mistake was to not see that the intersection of $\bar X$ with $H$ is not the reduced point $s$ but $s$ with a nilpotent structure.
Edit: detailed calculation
Indeed the intersection of $\bar X$ with the line at infinity $H$ is best computed in the affine plane $\mathbb A^2_{x,z}=Spec (k[x,z])$ (=the points of $\mathbb P^{2}$ where we may choose $y=1$), whose coordinates are $(x,z)=[x:1:z]$.
The point $s$ then has coordinates $x=0,z=0$, $\bar X$ has equation $z^2=x^3$, $H$ has equation $z=0$ and the ideal in $k[x,z]$ of the intersection $\bar X \cap H$ is $(z^2-x^3,z)=(z,x^3)$.
So the intersection $\bar X \cap H$ is the affine subscheme $Spec (k[x,z]/(z,x^3)) \subset \mathbb A^2_{x,z}$, which is clearly isomorphic to $A=Spec(k[x]/(x^3)$, hence non reduced and thus singular.
Geometrically $\bar X$ is a cusp cut by every projective line in three points. The line at infinity however cuts it in what was classically known as a "triple point" before the introduction of schemes.
Second edit
Let me add a few words to the second sentence of my answer in order to address Daniel's comment.
The problem is local at $s$, so we may assume that $s\in X$ is a singularity and we must show that $X\cap H $ is singular at $s$ too.
For simplicity assume that $X$ is a hypersurface.
It thus has equation $f(x_1,...,x_n)=0$, with $f(x_1,...,x_n)=q_2(x_1,...,x_n)+q_3(x_1,...,x_n)+...$ where $q_i(x_1,...,x_n)$ is homogeneous of degree $i$.
The crucial point is that there is no linear term $q_1$: this is equivalent to $X$ being singular at $s$.
If the hyperplane $H$ has equation $x_n=l(x_1,...,x_{n-1})$ ( $l$ linear) the intersection $X\cap H$ is given by $q_2(x_1,...,l(x_1,...,x_{n-1})+q_3(x_1,...,l(x_1,...,x_{n-1}))+...=0$ in the affine coordinates $x_1,...,x_{n-1}$ and is thus also singular since it begins with a quadratic term.