This is not true. For example, let $X = \ell^1$. Then $X^\ast = \ell^\infty$. Fix a free ultrafilter $\mathcal{U}$ on $\mathbb{N}$. Then $P: \ell^\infty \to \ell^\infty$, $P((a_n)_n) = (\lim_{k \to \mathcal{U}} a_k) \cdot (1)_n$ is a bounded linear projection onto the $1$-dimensional subspace $\text{span}\{(1)_n\}$ of $X^\ast$, which is clearly $\sigma(X^\ast, X)$-closed. However, $\text{ker}(P)$ contains $c_0$, which is $\sigma(X^\ast, X)$-dense, but $\text{ker}(P) \neq X^\ast$, so $\text{ker}(P)$ is not $\sigma(X^\ast, X)$-closed. Hence, $P$ is not $\sigma(X^\ast, X)$-continuous.

This is not true. For example, let $X = \ell^1$. Then $X^\ast = \ell^\infty$. Fix a free ultrafilter $\mathcal{U}$ on $\mathbb{N}$. Then $P: \ell^\infty \to \ell^\infty$, $P((a_n)_n) = (\lim_{k \to \mathcal{U}} a_k) \cdot (1)_n$ is a bounded linear projection onto the $1$-dimensional subspace $\text{span}\{(1)_n\}$ of $X^\ast$, which is clearly $\sigma(X^\ast, X)$-closed. However, $\text{ker}(P)$ contains $c_0$, which is $\sigma(X^\ast, X)$-dense, but $\text{ker}(P) \neq X^\ast$, so $\text{ker}(P)$ is not $\sigma(X^\ast, X)$-closed. Hence, $P$ is not $\sigma(X^\ast, X)$-continuous.

Let me provide a more general answer: a counterexample $P$ exists iff $X$ is non-reflexive. Clearly, if $X$ is reflexive, any bounded linear $P$ is automatically $\sigma(X^\ast, X)$-continuous, so no counterexample can exist. Otherwise, assume $X$ is non-reflexive. Then there exists $\varphi \in X^{\ast\ast} \setminus X$. Since $\varphi \neq 0$, we may pick $v \in X^\ast$ s.t. $\varphi(v) = 1$. Then $P(w) = \varphi(w) \cdot v$ is a bounded linear projection onto $\text{span}\{v\}$, which, being one-dimensional, is $\sigma(X^\ast, X)$-closed. However, since $\varphi$ is not $\sigma(X^\ast, X)$-continuous, neither is $P$.