= Solution
Assume the Laplace integrals converge absolutely in a common right half-plane. The <Fubini theorem> and the substitution $t=u-v$ on $u\geq v\geq0$ give the <convolution theorem for Laplace transforms>:
$$
\begin{aligned}
\widehat h(p)&=\int_0^\infty\int_0^u e^{-pu}f(u-v)g(v)\,dv\,du\\
&=\int_0^\infty\int_0^\infty e^{-p(t+v)}f(t)g(v)\,dt\,dv
=\boxed{\widehat f(p)\widehat g(p)}.
\end{aligned}
$$
For the specified inverse-square-root function, the supplied identity says $f*f=\pi$ on $t>0$. Consequently $\widehat f(p)^2=\pi/p$. For real $p>0$ the defining integral is positive, fixing the positive square root; analytic continuation into $\operatorname{Re}p>0$ gives
$$
\boxed{\widehat f(p)=\sqrt\pi\,p^{-1/2},}
$$
with the branch positive on the positive real axis. The square identity alone would not select the sign.
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