= Solution
Under $\mathbb Q$, $X_T=x+W_T^{\mathbb Q}$ has density $\varphi_T(z-x)=(2\pi T)^{-1/2}\exp(-(z-x)^2/(2T))$. The reciprocal <Radon-Nikodym derivative> depends only on $X_T$:
$$
\frac{d\mathbb P}{d\mathbb Q}=\frac{\cosh X_T}{\cosh x}e^{-T/2}.
$$
Consequently the <probability density function> under the original measure is
$$
\boxed{p_T(x,z)=\frac{\cosh z}{\cosh x}\frac{e^{-T/2}}{\sqrt{2\pi T}}\exp\left(-\frac{(z-x)^2}{2T}\right),\qquad z\in\mathbb R.}
$$
It integrates to one because $\mathbb E_{\mathbb Q}\cosh X_T=e^{T/2}\cosh x$. Completing the square also gives the useful <mixture distribution> form
$$
p_T(x,z)=\frac{e^x}{2\cosh x}\varphi_T(z-x-T)+\frac{e^{-x}}{2\cosh x}\varphi_T(z-x+T).
$$
Thus the terminal law is a mixture of $N(x+T,T)$ and $N(x-T,T)$ with the displayed positive weights. The <diffusion with hyperbolic tangent drift> density also exhibits the <Doob h-transform> with $h(z)=\cosh z$.
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