This is a long and detailed hint which provides ideas, not just for
this particular problem, but also for other magic graph problems
where similar reasoning can be used.
Vertex Magic
Investigation
There are 8 vertices and 7 edges. Let's suppose the magic constant
is $h$ and this is the same at each vertex. The total of all the
numbers $1 + 2 + ... + 15 = 120$ but the numbers on the edges, say
$a,b,c,d,e,f,g$ are each counted twice so
$$(a+b+c+d+e+f+g)+120=8h.$$
So $a+b+c+d+e+f+g$ is a multiple of 8 and it is at least
$1+2+...+7=28$ so it must be $32$ or more. Hence $8h\geq 152$ and
$h\geq 19$.
It is easy to check, in a similar way, for the largest value that
$h$ can take. Now it is a matter of systematically checking the
possibilities for magic labellings with $h=19$ etc.
The techniques for finding the labellings are the same as used
for the
Caterpillar problem.
Edge Magic
Investigation
How do we discover what edge magic graphs there are? Here the three
order-3 vertices $a, b,c$ are counted three times and the sum of
all the numbers is again $120$.
$$2(a+b+c)+120=7k.$$
Hence $k$ is even and since $a+b+c\geq 6$ we have $7k\geq
120+12=132$ and so $k\geq 20$.
If $k=20$ then $a+b+c=10$ so this triple is $1+2+7$ or $1+3+6$ or
$1+4+5$ or $2+3+5$.
The possible edge sums for $k=20$ are:
$1+4+15$, $1+5+14$, $1+6+13$, $1+7+12$, $1+8+11$, $1+9+10$,
$2+3+15$, $2+4+14$, $2+5+13$, $2+6+12$, $2+7+11$, $2+8+10$,
$3+4+13$, $3+5+12$, $3+6+11$, $3+7+10$, $3+8+9$, $4+5+11$,
$4+6+10$, $4+7+9$, $5+6+9$, $5+7+8$.
It is easy to find the edge magic labelling for $k=20$.
Moreover, substituting $16-P$ for each of the P-labels giving a
magic total of $k$ will give the same numbers $1$ to $15$ and a
magic total of $48-k$. So for all the magic graphs you find with a
magic total of $20$ there are corresponding graphs with a magic
total of $28$.
It is left for you to find the edge magic labels for $k=20$ and for
$k=22$ (and correspondingly $k=26$) andfor $k=24$. Remember $k$
must be even and at least $20$ so these are the only
possibilities.