The dimension if the Dirac delta function is the inverse of the dimension of its argument. So if $x$ is a length then $\delta(x)$ has the dimension of inverse length.

In your example, the momentum eigenstate in position representation has the wavefunction $$ \psi_p(x) = A \mathrm{e}^{i p x / \hbar} $$ The interpretation of the wavefunction is that $|\psi|^2$ is a probability density, which is dimension of 1/Length. Therefore, $A$ has dimension of $1/\sqrt{\mathrm{Length}}$.

As you say, calculating the scalar product of two momentum eigenfunction with eigenvalues $p_1$ and $p_2$ gives $$ \int_{-\infty}^\infty \mathrm{d}x \, \psi_{p_1}^*(x) \psi_{p_2}(x) = |A|^2 \int_{-\infty}^\infty \mathrm{d}x \mathrm{e}^{i(p_2-p_1)x/\hbar} = |A|^2 2\pi \hbar \delta(p_2 - p_1) $$ The left-hand side is dimensionless, and therefore so is the right-hand side. $\hbar \delta(p_2-p_1)$ has dimension of Length, so again we get that the dimension of $A$ is $1/\sqrt{\mathrm{Length}}$.

The dimension of the Dirac delta function is the inverse of the dimension of its argument. So if $x$ is a length then $\delta(x)$ has the dimension of inverse length.

In your example, the momentum eigenstate in position representation has the wavefunction $$ \psi_p(x) = A \mathrm{e}^{i p x / \hbar} $$ The interpretation of the wavefunction is that $|\psi|^2$ is a probability density, which is dimension of 1/Length. Therefore, $A$ has dimension of $1/\sqrt{\mathrm{Length}}$.

As you say, calculating the scalar product of two momentum eigenfunction with eigenvalues $p_1$ and $p_2$ gives $$ \int_{-\infty}^\infty \mathrm{d}x \, \psi_{p_1}^*(x) \psi_{p_2}(x) = |A|^2 \int_{-\infty}^\infty \mathrm{d}x \mathrm{e}^{i(p_2-p_1)x/\hbar} = |A|^2 2\pi \hbar \delta(p_2 - p_1) $$ The left-hand side is dimensionless, and therefore so is the right-hand side. $\hbar \delta(p_2-p_1)$ has dimension of Length, so again we get that the dimension of $A$ is $1/\sqrt{\mathrm{Length}}$.