Identify the relevant right-angled triangle.

Since D is directly above A, $\triangle DAB$ is a right-angled triangle with $\angle DAB = 90^\circ$.

We are given $AD = p$ and $\angle DBA = y$. We want to find $AB$.
In $\triangle DAB$:

The length of AB in terms of $p$ and $y$ is $\boxed{AB = \frac{p}{\tan y}}$.

We are given that A, B, C lie in the same horizontal plane, and $AB = AC$. This means $\triangle ABC$ is an isosceles triangle.
We are also given $\angle ABC = x$. Since $AB=AC$, then $\angle ACB = x$.
We are given $2AD = BC$. Since $AD = p$, then $BC = 2p$.

Draw a perpendicular line from A to BC, let the intersection point be M. Since $\triangle ABC$ is isosceles with $AB=AC$, the altitude AM bisects BC.
Therefore, $BM = MC = \frac{1}{2} BC = \frac{1}{2}(2p) = p$.

$\triangle AMB$ is a right-angled triangle with $\angle AMB = 90^\circ$.
In $\triangle AMB$:

From 8.1, we found $AB = \frac{p}{\tan y}$.
From Step 3, we found $AB = \frac{p}{\cos x}$.
Equating these two expressions for $AB$:

Divide both sides by $p$ (since $p \ne 0$):

Therefore, $\boxed{\cos x = \tan y}$.

We have $\cos x = \tan y$.

Given $x = 60^\circ$.

We know that $\cos 60^\circ = \frac{1}{2}$.

Rounding to one decimal place:

The size of $y$ is $\boxed{26.6^\circ}$.