= Solution
Construct a <mean-square derivative of a Gaussian process> on the original <probability space> before choosing its <continuous modification>. The real <covariance function> satisfies $K(h)=K(-h)$, because <covariance> is symmetric. Three-times differentiability implies that $K''$ is continuous everywhere and differentiable at zero. Since $K''$ is an <even function>,
$$
K'''(0)=0,\qquad K''(h)-K''(0)=o(|h|)\quad(h\to0).
$$
This uses differentiability of $K''$ at zero, not a bound on $K'''$ throughout a neighborhood.
Let $D_hX(t)=(X(t+h)-X(t))/h$ for $h\ne0$. Twice applying the <fundamental theorem of calculus> to the <covariance function> gives
$$
\mathbb E[D_hX(t)D_kX(s)]=-\frac1{hk}\int_0^h\int_0^k K''(t-s+u-v)\,dv\,du.
$$
Oriented integrals make this identity valid for negative as well as positive $h,k$. For a fixed $t$, the right side tends to $-K''(0)$ as $h,k\to0$. In particular,
$$
\mathbb E|D_hX(t)-D_kX(t)|^2\longrightarrow0.
$$
By completeness of <L2 space>, the <mean-square derivative of a Gaussian process>
$$
Y(t)=L^2\!\operatorname{-lim}_{h\to0}D_hX(t)
$$
exists. Finite vectors consisting of values of $X$ and $Y$ have a centered <multivariate normal distribution>: approximate the $Y$ coordinates by difference quotients and use <mean-square convergence> and <characteristic functions>. Thus $Y$ is a centered <Gaussian process> jointly constructed with $X$, with
$$
\mathbb E[Y(t)Y(s)]=-K''(t-s),\qquad \mathbb E[X(t)Y(s)]=-K'(t-s).
$$
In particular $-K''(0)\geq0$, and $-K''$ is a <positive-semidefinite kernel> because it is an actual <covariance function>. Merely constructing an unrelated <Gaussian process> with that <covariance function> would not yet construct a <modification of a stochastic process> of $X$.
The increment <variance> of $Y$ is
$$
\mathbb E|Y(t)-Y(s)|^2=2\bigl(K''(t-s)-K''(0)\bigr).
$$
For small $|t-s|$, differentiability of $K''$ at zero bounds this by $C|t-s|$. For the remaining distances in $[0,1]$, continuity of $K''$ enlarges $C$ to give the same bound. The fourth moment of a centered <normal distribution> is three times its <variance> squared, so
$$
\mathbb E|Y(t)-Y(s)|^4=3\bigl(\mathbb E|Y(t)-Y(s)|^2\bigr)^2\leq C_4|t-s|^2.
$$
The one-dimensional <Kolmogorov continuity theorem> states that $\mathbb E|U(t)-U(s)|^p\leq C|t-s|^{1+\eta}$ with $p,\eta>0$ gives a <continuous modification>, with sample-path <Hölder continuity> of every order less than $\eta/p$. Apply it here with $p=4,\eta=1$. We obtain a <continuous modification> $\widetilde Y$ on $[0,1]$. On its null exceptional event, set the entire path equal to zero.
Define, on this same <probability space>,
$$
\widetilde X(t)=X(0)+\int_0^t\widetilde Y(s)\,ds\qquad(0\leq t\leq1).
$$
The path integral exists because its integrand is a <continuous function>. It agrees in <L2 space> with the integral of $Y$: <Riemann sums> converge in <L2 space> because $Y$ is mean-square continuous; sums of $\widetilde Y$ have the same values almost surely at each finite collection of mesh points and converge pathwise to the integral. These two limits therefore agree almost surely for each fixed $t$. Equivalently, the integral is the <Bochner integral> of the <L2 space>-valued map $s\mapsto Y(s)$.
To verify the <modification of a stochastic process> property directly, put $A_t=X(t)-X(0)$ and $I_t=\int_0^tY(s)\,ds$. The preceding <covariance> identities and the <fundamental theorem of calculus> give
$$
\begin{aligned}
\mathbb E A_t^2&=2(K(0)-K(t)),\\
\mathbb E[A_tI_t]&=\int_0^t\bigl[-K'(t-s)-K'(s)\bigr]\,ds=2(K(0)-K(t)),\\
\mathbb E I_t^2&=-\int_0^t\int_0^tK''(s-u)\,ds\,du=2(K(0)-K(t)).
\end{aligned}
$$
Therefore $\mathbb E(A_t-I_t)^2=0$ and $\widetilde X(t)=X(t)$ almost surely for every fixed $t$. This is also the <mean-square fundamental theorem of calculus> in this stationary setting. If a version on all of $\mathbb R$ is desired, retain the original $X(t)$ outside $[0,1]$; equality at every fixed parameter still holds.
The ordinary <fundamental theorem of calculus> now gives the simultaneous pathwise conclusion
$$
\boxed{\widetilde X\in C^1([0,1])\text{ almost surely},\qquad \widetilde X'(t)=\widetilde Y(t)\quad(0<t<1).}
$$
Thus the requested <differentiable modification of a stationary Gaussian process> exists. Integrability of $K$ is compatible with the hypotheses but is unnecessary for this direct argument. It also avoids any assumption that a third absolute moment of a spectral density follows from three-times differentiability.
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