1) The Laplace transform of f(t) is $\overline{f}(p)=\frac{1}{p}$ when $f(t)=1$
2) The Laplace transform of $f(at)$ is $\frac{1}{a}\overline{f}(\frac{p}{a})$
3) The Laplace transform of the convolution $h(t)= f(t) \ast g(t)$ is $\mathcal{L}_{f \ast g} = \mathcal{L}_f \mathcal{L}_g$
4) You are given that if $f(t)=t^{-3/2}e^{-1/t}$ then $\overline{f}(p)=\pi^{1/2}e^{-2p^{1/2}}$
Now if the laplace transform of some function $\theta(x,t)$ is $\overline{\theta}(x,t)=p^{-1}e^{-xp^{1/2}}$. Then deduce that
$\theta(x,t)=erfc\Big(\frac{x}{2\sqrt(t)}\Big)$.
I can see that this of the form $\mathcal{L}_{f \ast g} = \mathcal{L}_f \mathcal{L}_g$ and so $f(t)=1$. But I cannot see how to use parts 2) and 4) to invert $\mathcal{L}_g$
Essentially, you want to evaluate
$$\int_0^t dt' \, t'^{-3/2} \, e^{-1/t'} $$
Sub $t'=1/u$ and get
$$\int_{1/t}^{\infty} du \, u^{-1/2} \, e^{-u} = 2 \int_{1/\sqrt{t}}^{\infty} dv \, e^{-v^2} = \sqrt{\pi} \operatorname{erfc}{\left (\frac1{\sqrt{t}} \right )}$$
Now you just need to scale the result to include the factor of $x$ and shuffle some constants around, and you are done.
ADDENDUM
Note that our result implies
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dp \, \frac1{p} e^{-2 \sqrt{p}} e^{p t} = \operatorname{erfc}{\left (\frac1{\sqrt{t}} \right )}$$
Let $p = 4 p'/x^2 $, then
$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} dp \frac1{p} \, e^{-x \sqrt{p}} e^{p t} &= \frac1{i 2 \pi} \int_{c'-i \infty}^{c'+i \infty} dp' \, \frac1{p'} e^{-2 \sqrt{p'}} e^{p' (4 t/x^2)}\\ &= \operatorname{erfc}{\left (\frac{x}{2\sqrt{t}} \right )} \end{align}$$