Well done all of you who sent in good solutions to this problem!
Many different methods were used. Saul Foresta and Julia Collins
found and used the exact value of $\tan 22.5$ degrees which they
found using trig formulae; Andre Lazanu used similar triangles;
Kamen Marinov used the Sine Rule and Pythagoras Theorem; Hyeyoun
Chung, Arun Ayer, Ngoc Tran and Yatir Halevi used the angle
bisector theorem ($BA$:$AO$=$BN$:$NO$); Robert Goudie used an
ingenious construction; and Dorothy Winn used several applications
of the Sine Rule.

## Robert Goudie's solution.

If you take triangle $AON$ and double its size you form triangle
$ACR$ since $AC$ is double $AO$.

So we now have two similar triangles $AON$ and $ACR$ and it is
clearly the case that $CR$ is double $ON$ so $CR = 48$.

Since $AR$ is the bisector of angle $OAB$ we know that angle
$CAR = 22.5^{\circ}$ . Therefore angle $CRA = 67.5^{\circ}$.

Angle $RCP$ must be $45^{\circ}$since angle $ACB$ is
$45^{\circ}$ and angle $ACR$ is $90^{\circ}$. This implies that
angle $CPR$ is $67.5^{\circ}$ since $180 - 45 - 67.5 = 67.5$.

Therefore triangle $PCR$ is isosceles and so $PC = CR$ and hence
$PC = 48$ units.

## Andre Lazanu's solution using similar triangles

I know that in a square the diagonals are also bisectors (in
fact, this is true for the rhombus, and the square is a particular
case of rhombus), so angle $BAC$ and angle $DAC$ have both the same
measure: $45^{\circ}$. The angles formed by the bisector of angle
$BAC$ with the sides of the angle have the following measure:
$22^{\circ}30`$.

I noted with $L$ the side of the square, using the Pythagorean
Theorem, $AC = L\sqrt{2}$

Now, I observe that $AO$ is half $AC$: $AO = (L\sqrt{2})/2$

Triangles $AON$ and $ABP$ are similar, because each one has a
right angle, and there is another pair of equal angles (the ones
formed by the bisector). So, I have:

$$\frac{AO}{AB} = \frac{AN}{AP} = \frac{ON}{BP}$$

So, I obtain the following proportion:

$(L\sqrt{2}/2) / L = 24/BP$

$BP = 48 / \sqrt{2} = 24\sqrt{2}$ (units)

Now, I see that angle $BPA$ has $67^{\circ}30`$ and angle $BNP$
has the same measure, so triangle $BNP$ is isosceles with $BP =
BN$.

Here I observe that triangles $ABN$ and $ACP$ are similar,
because their angles are respectively equal. So

$$\frac{AB}{AC} = \frac{BN}{PC} = \frac{AN}{AP}$$

As $AB / AC = 1 / \sqrt{2}$ it follows that $PC =
\sqrt{2}BN$.

So

$PC = \sqrt{2}BN = \sqrt{2}BP = 48$.